Merge branch 'master' of github.com:youngyangyang04/leetcode-master

problems/0018.四数之和.md

@@ -151,6 +151,96 @@ if (nums[k] + nums[i] > target && nums[i] >= 0) {
 
 ## 其他语言版本
 
+### C:
+
+```C
+/* qsort */
+static int cmp(const void* arg1, const void* arg2) {
+    int a = *(int *)arg1;
+    int b = *(int *)arg2;
+    return (a > b);
+}
+
+int** fourSum(int* nums, int numsSize, int target, int* returnSize, int** returnColumnSizes) {
+
+    /* 对nums数组进行排序 */
+    qsort(nums, numsSize, sizeof(int), cmp);
+
+    int **res = (int **)malloc(sizeof(int *) * 40000);
+    int index = 0;
+
+    /* k */
+    for (int k = 0; k < numsSize - 3; k++) { /* 第一级 */
+
+        /* k剪枝 */
+        if ((nums[k] > target) && (nums[k] >= 0)) {
+            break;
+        }
+        /* k去重 */
+        if ((k > 0) && (nums[k] == nums[k - 1])) {
+            continue;
+        }
+
+        /* i */
+        for (int i = k + 1; i < numsSize - 2; i++) { /* 第二级 */
+
+            /* i剪枝 */
+            if ((nums[k] + nums[i] > target) && (nums[i] >= 0)) {
+                break;
+            }
+            /* i去重 */
+            if ((i > (k + 1)) && (nums[i] == nums[i - 1])) {
+                continue;
+            }
+
+            /* left and right */
+            int left = i + 1;
+            int right = numsSize - 1;
+
+            while (left < right) {
+
+                /* 防止大数溢出 */
+                long long val = (long long)nums[k] + nums[i] + nums[left] + nums[right];
+                if (val > target) {
+                    right--;
+                } else if (val < target) {
+                    left++;
+                } else {
+                    int *res_tmp = (int *)malloc(sizeof(int) * 4);
+                    res_tmp[0] = nums[k];
+                    res_tmp[1] = nums[i];
+                    res_tmp[2] = nums[left];
+                    res_tmp[3] = nums[right];
+                    res[index++] = res_tmp;
+                    
+                    /* right去重 */
+                    while ((right > left) && (nums[right] == nums[right - 1])) {
+                        right--;
+                    }
+                    /* left去重 */
+                    while ((left < right) && (nums[left] == nums[left + 1])) {
+                        left++;
+                    }
+
+                    /* 更新right与left */
+                    left++, right--;
+                }
+            }
+        }
+    }
+
+    /* 返回值处理 */
+    *returnSize = index;
+
+    int *column = (int *)malloc(sizeof(int) * index);
+    for (int i = 0; i < index; i++) {
+        column[i] = 4;
+    }
+    *returnColumnSizes = column;
+    return res;
+}
+```
+
 ### Java：
 
 ```Java

problems/0063.不同路径II.md

@@ -550,6 +550,27 @@ function uniquePathsWithObstacles(obstacleGrid: number[][]): number {
 };
 ```
 
+// 版本二: dp改為使用一維陣列，從終點開始遍歷
+```typescript
+function uniquePathsWithObstacles(obstacleGrid: number[][]): number {
+  const m = obstacleGrid.length;
+  const n = obstacleGrid[0].length;
+  
+  const dp: number[] = new Array(n).fill(0);
+  dp[n - 1] = 1;
+
+  // 由下而上，右而左進行遍歷
+  for (let i = m - 1; i >= 0; i--) {
+    for (let j = n - 1; j >= 0; j--) {
+      if (obstacleGrid[i][j] === 1) dp[j] = 0;
+      else dp[j] = dp[j] + (dp[j + 1] || 0); 
+    }
+  }
+
+  return dp[0];
+};
+```
+
 ### Rust
 
 ```Rust

problems/1047.删除字符串中的所有相邻重复项.md

@@ -241,6 +241,33 @@ class Solution:
 
 ### Go：
 
+使用栈
+```go
+func removeDuplicates(s string) string {
+    stack := make([]rune, 0)
+    for _, val := range s {
+        if len(stack) == 0 || val != stack[len(stack)-1] {
+            stack = append(stack, val)
+        } else {
+            stack = stack[:len(stack)-1]
+        }
+    }
+    var res []rune
+    for len(stack) != 0 { // 将栈中元素放到result字符串汇总
+        res = append(res, stack[len(stack)-1])
+        stack = stack[:len(stack)-1]
+    }
+    // 此时字符串需要反转一下
+    l, r := 0, len(res)-1
+    for l < r {
+        res[l], res[r] = res[r], res[l]
+        l++
+        r--
+    }
+    return string(res)
+}
+```
+拿字符串直接作为栈，省去了栈还要转为字符串的操作
 ```go
 func removeDuplicates(s string) string {
     var stack []byte

problems/kamacoder/0054.替换数字.md

@@ -362,10 +362,43 @@ main();
 
 ### Scala:
 
-
 ### PHP：
 
+```php
+<?php
+// 标准输入
+$s = trim(fgets(STDIN));
+$oldLen = strlen($s);
+$count = 0;
+for ($i = 0; $i < $oldLen; $i++) {
+    if (is_numeric($s[$i])) {
+      $count++;
+    }
+}
+
+// 扩充字符串
+$s = str_pad($s, $oldLen + $count * 5);
+$newLen = strlen($s);
+while($oldLen >= 0) {
+    if (is_numeric($s[$oldLen])) {
+        $s[$newLen--] = 'r';
+        $s[$newLen--] = 'e';
+        $s[$newLen--] = 'b';
+        $s[$newLen--] = 'm';
+        $s[$newLen--] = 'u';
+        $s[$newLen--] = 'n';
+    } else {
+        $s[$newLen--] = $s[$oldLen];
+    }
+    $oldLen--;
+}
+
+echo $s;
+?>
+```
+
+
+
 
 ### Rust:
 
-

problems/kamacoder/0055.右旋字符串.md

@@ -300,33 +300,50 @@ int main()
 ```javascript
 // JS中字符串内不可单独修改
 
-// 右旋转
+const readline = require('readline')
-function reverseLeftWords(s, k) {
+
-  const reverse = (sList, start, end) => {
+const rl = readline.createInterface({
-    for (let i = start, j = end; i < j; i++, j--) {
+    input: process.stdin,
-      [sList[i], sList[j]] = [sList[j], sList[i]];
+    output: process.stdout
-    }
+})
-  }
+
-  const sList = Array.from(s);
+const inputs = []; // 存储输入
-  reverse(sList, 0, sList.length - k - 1);
+
-  reverse(sList, sList.length - k, sList.length - 1);
+rl.on('line', function(data) {
-  reverse(sList, 0, sList.length - 1);
+    inputs.push(data);
-  return sList.join('');
+
+}).on('close', function() {
+    const res = deal(inputs);
+    // 打印结果
+    console.log(res);
+})
+
+// 对传入的数据进行处理
+function deal(inputs) {
+    let [k, s] = inputs;
+    const len = s.length - 1;
+    k = parseInt(k);
+    str = s.split('');
+    
+    str = reverseStr(str, 0, len - k)
+    str = reverseStr(str, len - k + 1, len)
+    str = reverseStr(str, 0, len)
+
+    return str.join('');
 }
 
-// 左旋转
+// 根据提供的范围进行翻转
-var reverseLeftWords = function(s, n) {
+function reverseStr(s, start, end) {
-  const reverse = (sList, start, end) => {
+    
-    for (let i = start, j = end; i < j; i++, j--) {
+    while (start < end) {
-      [sList[i], sList[j]] = [sList[j], sList[i]];
+        [s[start], s[end]] = [s[end], s[start]]
+        
+        start++;
+        end--;
     }
-  }
+
-  const sList = s.split('');
+    return s;
-  reverse(sList, 0, n - 1);
+}
-  reverse(sList, n, sList.length - 1);
-  reverse(sList, 0, sList.length - 1);
-  return sList.join('');
-};
 ```
 
 ### TypeScript：
@@ -338,10 +355,35 @@ var reverseLeftWords = function(s, n) {
 
 ### PHP：
 
+```php
+<?php
+// 反转函数
+function reverse(&$s, $start, $end) {
+    for ($i = $start, $j = $end; $i < $j; $i++, $j--) {
+        $tmp = $s[$i];
+        $s[$i] = $s[$j];
+        $s[$j] = $tmp;
+    }
+}
+    
+// 标准输入：读取右旋转位数和字符串    
+$n = trim(fgets(STDIN));
+$s = trim(fgets(STDIN));
+// 字符串长度
+$len = strlen($s);
+// 先部分反转
+reverse($s, $len - $n, $len - 1);
+reverse($s, 0, $len - $n - 1);
+// 再整体反转
+reverse($s, 0, $len - 1);
+
+echo $s;
+?>
+```
+
 
 ### Scala:
 
 
 ### Rust:
 
-

problems/kamacoder/0101.孤岛的总面积.md

@@ -186,7 +186,56 @@ int main() {
 ### Java 
 
 ### Python
+```python
+from collections import deque
 
+# 处理输入
+n, m = list(map(int, input().strip().split()))
+g = []
+for _ in range(n):
+    row = list(map(int, input().strip().split()))
+    g.append(row)
+
+# 定义四个方向、孤岛面积（遍历完边缘后会被重置）
+directions = [[0,1], [1,0], [-1,0], [0,-1]]
+count = 0
+
+# 广搜
+def bfs(r, c):
+    global count
+    q = deque()
+    q.append((r, c))
+    g[r][c] = 0
+    count += 1
+
+    while q:
+        r, c = q.popleft()
+        for di in directions:
+            next_r = r + di[0]
+            next_c = c + di[1]
+            if next_c < 0 or next_c >= m or next_r < 0 or next_r >= n:
+                continue
+            if g[next_r][next_c] == 1:
+                q.append((next_r, next_c))
+                g[next_r][next_c] = 0
+                count += 1
+
+
+for i in range(n):
+    if g[i][0] == 1: bfs(i, 0)
+    if g[i][m-1] == 1: bfs(i, m-1)
+
+for i in range(m):
+    if g[0][i] == 1: bfs(0, i)
+    if g[n-1][i] == 1: bfs(n-1, i)
+
+count = 0
+for i in range(n):
+    for j in range(m):
+        if g[i][j] == 1: bfs(i, j)
+
+print(count)
+```
 ### Go
 
 ### Rust

problems/qita/join.md

@@ -188,7 +188,7 @@ python代码
 大家提交代码的热情太高了，我有时候根本处理不过来，但我必须当天处理完，否则第二天代码冲突会越来越多。
 <div align="center"><img src='https://code-thinking-1253855093.file.myqcloud.com/pics/20210514091457392.png' width=500 alt=''></img></div>
 
-一天晚分别有两位录友提交了 30多道 java代码，全部冲突，解决冲突处理的我脖子疼[哭]
+一天晚上分别有两位录友提交了 30多道 java代码，全部冲突，解决冲突处理的我脖子疼[哭]
 
 那么在处理冲突的时候 保留谁的代码，删点谁的代码呢？