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feat: add solutions to lc problem: No.1106
No.1106.Parsing A Boolean Expression
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‎solution/1100-1199/1106.Parsing A Boolean Expression/README.md

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Original file line numberDiff line numberDiff line change
@@ -58,22 +58,153 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：栈**
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对于这种表达式解析问题，我们可以使用栈来辅助解决。
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从左到右遍历表达式 `expression`，对于遍历到的每个字符 $c$：
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- 如果 $c$ 是 `"tf!&|"` 中的一个，我们直接将其入栈；
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- 如果 $c$ 是右括号 `')'`，我们将栈中元素依次出栈，直到遇到操作符 `'!'` 或 `'&'` 或 `'|'`。过程中我们用变量 $t$ 和 $f$ 记录出栈字符中 `'t'` 和 `'f'` 的个数。最后根据出栈字符的个数和操作符计算得到新的字符 `'t'` 或 `'f'`，并将其入栈。
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遍历完表达式 `expression` 后，栈中只剩下一个字符，如果是 `'t'`，返回 `true`，否则返回 `false`。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def parseBoolExpr(self, expression: str) -> bool:
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stk = []
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for c in expression:
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if c in 'tf!&|':
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stk.append(c)
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elif c == ')':
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t = f = 0
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while stk[-1] in 'tf':
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t += stk[-1] == 't'
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f += stk[-1] == 'f'
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stk.pop()
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match stk.pop():
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case '!':
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c = 't' if f else 'f'
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case '&':
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c = 'f' if f else 't'
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case '|':
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c = 't' if t else 'f'
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stk.append(c)
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return stk[0] == 't'
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public boolean parseBoolExpr(String expression) {
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Deque<Character> stk = new ArrayDeque<>();
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for (int i = 0; i < expression.length(); ++i) {
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char c = expression.charAt(i);
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if (c != '(' && c != ')' && c != ',') {
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stk.push(c);
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} else if (c == ')') {
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int t = 0, f = 0;
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while (stk.peek() == 't' || stk.peek() == 'f') {
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t += stk.peek() == 't' ? 1 : 0;
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f += stk.peek() == 'f' ? 1 : 0;
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stk.pop();
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}
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char op = stk.pop();
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if (op == '!') {
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c = f > 0 ? 't' : 'f';
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} else if (op == '&') {
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c = f > 0 ? 'f' : 't';
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} else {
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c = t > 0 ? 't' : 'f';
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}
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stk.push(c);
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}
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}
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return stk.peek() == 't';
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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bool parseBoolExpr(string expression) {
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stack<char> stk;
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for (char c : expression) {
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if (c != '(' && c != ')' && c != ',') stk.push(c);
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else if (c == ')') {
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int t = 0, f = 0;
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while (stk.top() == 't' || stk.top() == 'f') {
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t += stk.top() == 't';
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f += stk.top() == 'f';
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stk.pop();
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}
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char op = stk.top();
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stk.pop();
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if (op == '!') c = f ? 't' : 'f';
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if (op == '&') c = f ? 'f' : 't';
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if (op == '|') c = t ? 't' : 'f';
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stk.push(c);
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}
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}
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return stk.top() == 't';
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}
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};
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```
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### **Go**
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```go
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func parseBoolExpr(expression string) bool {
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stk := []rune{}
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for _, c := range expression {
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if c != '(' && c != ')' && c != ',' {
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stk = append(stk, c)
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} else if c == ')' {
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var t, f int
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for stk[len(stk)-1] == 't' || stk[len(stk)-1] == 'f' {
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if stk[len(stk)-1] == 't' {
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t++
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} else {
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f++
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}
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stk = stk[:len(stk)-1]
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}
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op := stk[len(stk)-1]
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stk = stk[:len(stk)-1]
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c = 'f'
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switch op {
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case '!':
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if f > 0 {
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c = 't'
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}
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case '&':
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if f == 0 {
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c = 't'
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}
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case '|':
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if t > 0 {
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c = 't'
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}
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}
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stk = append(stk, c)
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}
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}
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return stk[0] == 't'
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}
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```
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### **...**

‎solution/1100-1199/1106.Parsing A Boolean Expression/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -63,13 +63,131 @@ Then, evaluate !(f) --&gt; NOT false --&gt; true. We return true.
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### **Python3**
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```python
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class Solution:
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def parseBoolExpr(self, expression: str) -> bool:
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stk = []
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for c in expression:
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if c in 'tf!&|':
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stk.append(c)
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elif c == ')':
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t = f = 0
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while stk[-1] in 'tf':
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t += stk[-1] == 't'
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f += stk[-1] == 'f'
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stk.pop()
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match stk.pop():
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case '!':
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c = 't' if f else 'f'
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case '&':
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c = 'f' if f else 't'
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case '|':
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c = 't' if t else 'f'
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stk.append(c)
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return stk[0] == 't'
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```
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### **Java**
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```java
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class Solution {
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public boolean parseBoolExpr(String expression) {
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Deque<Character> stk = new ArrayDeque<>();
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for (int i = 0; i < expression.length(); ++i) {
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char c = expression.charAt(i);
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if (c != '(' && c != ')' && c != ',') {
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stk.push(c);
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} else if (c == ')') {
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int t = 0, f = 0;
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while (stk.peek() == 't' || stk.peek() == 'f') {
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t += stk.peek() == 't' ? 1 : 0;
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f += stk.peek() == 'f' ? 1 : 0;
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stk.pop();
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}
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char op = stk.pop();
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if (op == '!') {
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c = f > 0 ? 't' : 'f';
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} else if (op == '&') {
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c = f > 0 ? 'f' : 't';
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} else {
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c = t > 0 ? 't' : 'f';
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}
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stk.push(c);
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}
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}
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return stk.peek() == 't';
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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bool parseBoolExpr(string expression) {
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stack<char> stk;
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for (char c : expression) {
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if (c != '(' && c != ')' && c != ',') stk.push(c);
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else if (c == ')') {
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int t = 0, f = 0;
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while (stk.top() == 't' || stk.top() == 'f') {
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t += stk.top() == 't';
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f += stk.top() == 'f';
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stk.pop();
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}
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char op = stk.top();
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stk.pop();
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if (op == '!') c = f ? 't' : 'f';
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if (op == '&') c = f ? 'f' : 't';
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if (op == '|') c = t ? 't' : 'f';
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stk.push(c);
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}
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}
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return stk.top() == 't';
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}
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};
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```
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### **Go**
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```go
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func parseBoolExpr(expression string) bool {
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stk := []rune{}
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for _, c := range expression {
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if c != '(' && c != ')' && c != ',' {
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stk = append(stk, c)
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} else if c == ')' {
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var t, f int
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for stk[len(stk)-1] == 't' || stk[len(stk)-1] == 'f' {
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if stk[len(stk)-1] == 't' {
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t++
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} else {
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f++
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}
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stk = stk[:len(stk)-1]
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}
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op := stk[len(stk)-1]
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stk = stk[:len(stk)-1]
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c = 'f'
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switch op {
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case '!':
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if f > 0 {
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c = 't'
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}
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case '&':
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if f == 0 {
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c = 't'
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}
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case '|':
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if t > 0 {
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c = 't'
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}
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}
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stk = append(stk, c)
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}
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}
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return stk[0] == 't'
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}
73191
```
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### **...**

‎solution/1100-1199/1106.Parsing A Boolean Expression/Solution.cpp

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,24 @@
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class Solution {
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public:
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bool parseBoolExpr(string expression) {
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stack<char> stk;
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for (char c : expression) {
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if (c != '(' && c != ')' && c != ',') stk.push(c);
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else if (c == ')') {
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int t = 0, f = 0;
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while (stk.top() == 't' || stk.top() == 'f') {
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t += stk.top() == 't';
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f += stk.top() == 'f';
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stk.pop();
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}
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char op = stk.top();
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stk.pop();
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if (op == '!') c = f ? 't' : 'f';
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if (op == '&') c = f ? 'f' : 't';
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if (op == '|') c = t ? 't' : 'f';
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stk.push(c);
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}
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}
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return stk.top() == 't';
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}
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};

‎solution/1100-1199/1106.Parsing A Boolean Expression/Solution.go

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func parseBoolExpr(expression string) bool {
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stk := []rune{}
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for _, c := range expression {
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if c != '(' && c != ')' && c != ',' {
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stk = append(stk, c)
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} else if c == ')' {
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var t, f int
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for stk[len(stk)-1] == 't' || stk[len(stk)-1] == 'f' {
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if stk[len(stk)-1] == 't' {
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t++
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} else {
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f++
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}
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stk = stk[:len(stk)-1]
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}
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op := stk[len(stk)-1]
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stk = stk[:len(stk)-1]
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c = 'f'
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switch op {
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case '!':
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if f > 0 {
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c = 't'
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}
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case '&':
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if f == 0 {
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c = 't'
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}
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case '|':
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if t > 0 {
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c = 't'
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}
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}
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stk = append(stk, c)
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}
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}
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return stk[0] == 't'
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}

‎solution/1100-1199/1106.Parsing A Boolean Expression/Solution.java

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class Solution {
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public boolean parseBoolExpr(String expression) {
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Deque<Character> stk = new ArrayDeque<>();
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for (int i = 0; i < expression.length(); ++i) {
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char c = expression.charAt(i);
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if (c != '(' && c != ')' && c != ',') {
7+
stk.push(c);
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} else if (c == ')') {
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int t = 0, f = 0;
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while (stk.peek() == 't' || stk.peek() == 'f') {
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t += stk.peek() == 't' ? 1 : 0;
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f += stk.peek() == 'f' ? 1 : 0;
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stk.pop();
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}
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char op = stk.pop();
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if (op == '!') {
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c = f > 0 ? 't' : 'f';
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} else if (op == '&') {
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c = f > 0 ? 'f' : 't';
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} else {
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c = t > 0 ? 't' : 'f';
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}
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stk.push(c);
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}
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}
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return stk.peek() == 't';
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}
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}

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