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feat: add solutions to lc problem: No.1017
No.1017.Convert to Base -2
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6 files changed

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‎solution/1000-1099/1017.Convert to Base -2/README.md

Lines changed: 93 additions & 1 deletion
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@@ -48,22 +48,114 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：模拟**
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我们可以判断 $n$ 从低位到高位的每一位，如果该位为 $1$，那么答案的该位为 $1$，否则为 $0$。如果该位为 $1$，那么我们需要将 $n$ 减去 $k$。接下来我们更新 $n = \lfloor n / 2 \rfloor$, $k = -k$。继续判断下一位。
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最后，我们将答案反转后返回即可。
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时间复杂度 $O(\log n)$，其中 $n$ 为给定的整数。忽略答案的空间消耗，空间复杂度 $O(1)$。
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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
58-
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class Solution:
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def baseNeg2(self, n: int) -> str:
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k = 1
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ans = []
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while n:
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if n % 2:
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ans.append('1')
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n -= k
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else:
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ans.append('0')
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n //= 2
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k *= -1
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return ''.join(ans[::-1]) or '0'
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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 String baseNeg2(int n) {
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if (n == 0) {
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return "0";
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}
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int k = 1;
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StringBuilder ans = new StringBuilder();
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while (n != 0) {
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if (n % 2 != 0) {
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ans.append(1);
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n -= k;
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} else {
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ans.append(0);
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}
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k *= -1;
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n /= 2;
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}
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return ans.reverse().toString();
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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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string baseNeg2(int n) {
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if (n == 0) {
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return "0";
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}
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int k = 1;
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string ans;
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while (n) {
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if (n % 2) {
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ans.push_back('1');
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n -= k;
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} else {
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ans.push_back('0');
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}
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k *= -1;
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n /= 2;
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}
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reverse(ans.begin(), ans.end());
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return ans;
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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 baseNeg2(n int) string {
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if n == 0 {
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return "0"
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}
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ans := []byte{}
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k := 1
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for n != 0 {
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if n%2 != 0 {
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ans = append(ans, '1')
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n -= k
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} else {
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ans = append(ans, '0')
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}
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k *= -1
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n /= 2
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}
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for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
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ans[i], ans[j] = ans[j], ans[i]
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}
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return string(ans)
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}
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```
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### **...**

‎solution/1000-1099/1017.Convert to Base -2/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -47,13 +47,97 @@
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### **Python3**
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```python
50-
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class Solution:
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def baseNeg2(self, n: int) -> str:
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k = 1
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ans = []
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while n:
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if n % 2:
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ans.append('1')
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n -= k
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else:
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ans.append('0')
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n //= 2
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k *= -1
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return ''.join(ans[::-1]) or '0'
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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 String baseNeg2(int n) {
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if (n == 0) {
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return "0";
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}
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int k = 1;
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StringBuilder ans = new StringBuilder();
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while (n != 0) {
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if (n % 2 != 0) {
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ans.append(1);
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n -= k;
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} else {
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ans.append(0);
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}
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k *= -1;
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n /= 2;
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}
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return ans.reverse().toString();
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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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string baseNeg2(int n) {
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if (n == 0) {
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return "0";
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}
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int k = 1;
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string ans;
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while (n) {
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if (n % 2) {
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ans.push_back('1');
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n -= k;
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} else {
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ans.push_back('0');
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}
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k *= -1;
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n /= 2;
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}
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reverse(ans.begin(), ans.end());
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return ans;
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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 baseNeg2(n int) string {
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if n == 0 {
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return "0"
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}
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ans := []byte{}
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k := 1
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for n != 0 {
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if n%2 != 0 {
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ans = append(ans, '1')
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n -= k
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} else {
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ans = append(ans, '0')
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}
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k *= -1
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n /= 2
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}
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for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
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ans[i], ans[j] = ans[j], ans[i]
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}
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return string(ans)
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}
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```
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### **...**

‎solution/1000-1099/1017.Convert to Base -2/Solution.cpp

Lines changed: 22 additions & 0 deletions
Original file line numberDiff line numberDiff line change
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class Solution {
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public:
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string baseNeg2(int n) {
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if (n == 0) {
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return "0";
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}
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int k = 1;
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string ans;
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while (n) {
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if (n % 2) {
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ans.push_back('1');
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n -= k;
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} else {
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ans.push_back('0');
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}
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k *= -1;
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n /= 2;
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}
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reverse(ans.begin(), ans.end());
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return ans;
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}
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};

‎solution/1000-1099/1017.Convert to Base -2/Solution.go

Lines changed: 21 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,21 @@
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func baseNeg2(n int) string {
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if n == 0 {
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return "0"
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}
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ans := []byte{}
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k := 1
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for n != 0 {
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if n%2 != 0 {
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ans = append(ans, '1')
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n -= k
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} else {
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ans = append(ans, '0')
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}
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k *= -1
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n /= 2
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}
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for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
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ans[i], ans[j] = ans[j], ans[i]
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}
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return string(ans)
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}

‎solution/1000-1099/1017.Convert to Base -2/Solution.java

Lines changed: 15 additions & 8 deletions
Original file line numberDiff line numberDiff line change
@@ -1,13 +1,20 @@
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class Solution {
2-
public String baseNeg2(int N) {
3-
if (N == 0) {
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public String baseNeg2(int n) {
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if (n == 0) {
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return "0";
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}
6-
StringBuilder sb = new StringBuilder();
7-
while (N != 0) {
8-
sb.append(N & 1);
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N = -(N >> 1);
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int k = 1;
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StringBuilder ans = new StringBuilder();
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while (n != 0) {
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if (n % 2 != 0) {
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ans.append(1);
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n -= k;
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} else {
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ans.append(0);
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}
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k *= -1;
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n /= 2;
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}
11-
return sb.reverse().toString();
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return ans.reverse().toString();
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}
13-
}
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}

‎solution/1000-1099/1017.Convert to Base -2/Solution.py

Lines changed: 13 additions & 0 deletions
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@@ -0,0 +1,13 @@
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class Solution:
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def baseNeg2(self, n: int) -> str:
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k = 1
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ans = []
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while n:
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if n % 2:
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ans.append('1')
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n -= k
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else:
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ans.append('0')
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n //= 2
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k *= -1
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return ''.join(ans[::-1]) or '0'

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