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feat: add solutions to lc problem: No.1003
No.1003.Check If Word Is Valid After Substitutions
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6 files changed

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‎solution/1000-1099/1003.Check If Word Is Valid After Substitutions/README.md

Lines changed: 84 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -56,22 +56,105 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：栈模拟**
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先判断字符串长度是否为 $3$ 的倍数，若不是直接返回 `false`。
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接下来我们使用栈模拟操作，遍历字符串 $s$ 的每个字符 $c$：
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若 $c$ 等于 `'c'`，且栈顶的两个元素分别为 `'a'` 和 `'b'`，则将栈顶的两个元素出栈；否则将 $c$ 入栈。
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最后判断栈是否为空，若为空则返回 `true`，否则返回 `false`。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串 $s$ 的长度。
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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 isValid(self, s: str) -> bool:
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if len(s) % 3:
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return False
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stk = []
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for c in s:
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if c == 'c' and len(stk) > 1 and stk[-2] == 'a' and stk[-1] == 'b':
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stk = stk[:-2]
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else:
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stk.append(c)
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return not stk
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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 isValid(String s) {
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if (s.length() % 3 > 0) {
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return false;
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}
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StringBuilder stk = new StringBuilder();
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for (char c : s.toCharArray()) {
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int n = stk.length();
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if (c == 'c' && n > 1 && stk.charAt(n - 2) == 'a' && stk.charAt(n - 1) == 'b') {
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stk.deleteCharAt(n - 1);
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stk.deleteCharAt(n - 2);
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} else {
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stk.append(c);
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}
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}
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return stk.length() == 0;
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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 isValid(string s) {
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if (s.size() % 3) {
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return false;
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}
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string stk;
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for (char c : s) {
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int n = stk.size();
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if (c == 'c' && n > 1 && stk[n - 2] == 'a' && stk[n - 1] == 'b') {
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stk.pop_back();
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stk.pop_back();
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} else {
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stk.push_back(c);
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}
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}
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return stk.empty();
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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 isValid(s string) bool {
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if len(s)%3 > 0 {
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return false
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}
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stk := []rune{}
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for _, c := range s {
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n := len(stk)
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if c == 'c' && n > 1 && stk[n-2] == 'a' && stk[n-1] == 'b' {
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stk = stk[:n-2]
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} else {
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stk = append(stk, c)
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}
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}
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return len(stk) == 0
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}
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```
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### **...**

‎solution/1000-1099/1003.Check If Word Is Valid After Substitutions/README_EN.md

Lines changed: 72 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -57,13 +57,84 @@ Thus, &quot;abcabcababcc&quot; is valid.
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### **Python3**
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```python
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class Solution:
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def isValid(self, s: str) -> bool:
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if len(s) % 3:
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return False
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stk = []
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for c in s:
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if c == 'c' and len(stk) > 1 and stk[-2] == 'a' and stk[-1] == 'b':
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stk = stk[:-2]
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else:
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stk.append(c)
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return not stk
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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 isValid(String s) {
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if (s.length() % 3 > 0) {
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return false;
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}
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StringBuilder stk = new StringBuilder();
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for (char c : s.toCharArray()) {
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int n = stk.length();
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if (c == 'c' && n > 1 && stk.charAt(n - 2) == 'a' && stk.charAt(n - 1) == 'b') {
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stk.deleteCharAt(n - 1);
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stk.deleteCharAt(n - 2);
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} else {
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stk.append(c);
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}
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}
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return stk.length() == 0;
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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 isValid(string s) {
102+
if (s.size() % 3) {
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return false;
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}
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string stk;
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for (char c : s) {
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int n = stk.size();
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if (c == 'c' && n > 1 && stk[n - 2] == 'a' && stk[n - 1] == 'b') {
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stk.pop_back();
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stk.pop_back();
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} else {
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stk.push_back(c);
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}
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}
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return stk.empty();
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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 isValid(s string) bool {
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if len(s)%3 > 0 {
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return false
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}
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stk := []rune{}
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for _, c := range s {
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n := len(stk)
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if c == 'c' && n > 1 && stk[n-2] == 'a' && stk[n-1] == 'b' {
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stk = stk[:n-2]
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} else {
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stk = append(stk, c)
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}
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}
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return len(stk) == 0
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}
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```
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### **...**

‎solution/1000-1099/1003.Check If Word Is Valid After Substitutions/Solution.cpp

Lines changed: 19 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,19 @@
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class Solution {
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public:
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bool isValid(string s) {
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if (s.size() % 3) {
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return false;
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}
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string stk;
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for (char c : s) {
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int n = stk.size();
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if (c == 'c' && n > 1 && stk[n - 2] == 'a' && stk[n - 1] == 'b') {
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stk.pop_back();
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stk.pop_back();
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} else {
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stk.push_back(c);
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}
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}
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return stk.empty();
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}
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};

‎solution/1000-1099/1003.Check If Word Is Valid After Substitutions/Solution.go

Lines changed: 15 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,15 @@
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func isValid(s string) bool {
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if len(s)%3 > 0 {
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return false
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}
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stk := []rune{}
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for _, c := range s {
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n := len(stk)
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if c == 'c' && n > 1 && stk[n-2] == 'a' && stk[n-1] == 'b' {
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stk = stk[:n-2]
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} else {
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stk = append(stk, c)
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}
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}
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return len(stk) == 0
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}

‎solution/1000-1099/1003.Check If Word Is Valid After Substitutions/Solution.java

Lines changed: 18 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,18 @@
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class Solution {
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public boolean isValid(String s) {
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if (s.length() % 3 > 0) {
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return false;
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}
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StringBuilder stk = new StringBuilder();
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for (char c : s.toCharArray()) {
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int n = stk.length();
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if (c == 'c' && n > 1 && stk.charAt(n - 2) == 'a' && stk.charAt(n - 1) == 'b') {
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stk.deleteCharAt(n - 1);
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stk.deleteCharAt(n - 2);
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} else {
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stk.append(c);
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}
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}
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return stk.length() == 0;
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}
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}

‎solution/1000-1099/1003.Check If Word Is Valid After Substitutions/Solution.py

Lines changed: 11 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,11 @@
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class Solution:
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def isValid(self, s: str) -> bool:
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if len(s) % 3:
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return False
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stk = []
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for c in s:
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if c == 'c' and len(stk) > 1 and stk[-2] == 'a' and stk[-1] == 'b':
8+
stk = stk[:-2]
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else:
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stk.append(c)
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return not stk

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