Merge branch 'youngyangyang04:master' into master

Author: WuTao1103

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/1020.飞地的数量.md

@@ -605,6 +605,63 @@ func bfs(grid [][]int, i, j int) {
 }
 ```
 
+### JavaScript
+
+```js
+/**
+ * @param {number[][]} grid
+ * @return {number}
+ */
+var numEnclaves = function (grid) {
+  let row = grid.length;
+  let col = grid[0].length;
+  let count = 0;
+
+  // Check the first and last row, if there is a 1, then change all the connected 1s to 0 and don't count them.
+  for (let j = 0; j < col; j++) {
+    if (grid[0][j] === 1) {
+      dfs(0, j, false);
+    }
+    if (grid[row - 1][j] === 1) {
+      dfs(row - 1, j, false);
+    }
+  }
+
+  // Check the first and last column, if there is a 1, then change all the connected 1s to 0 and don't count them.
+  for (let i = 0; i < row; i++) {
+    if (grid[i][0] === 1) {
+      dfs(i, 0, false);
+    }
+    if (grid[i][col - 1] === 1) {
+      dfs(i, col - 1, false);
+    }
+  }
+
+  // Check the rest of the grid, if there is a 1, then change all the connected 1s to 0 and count them.
+  for (let i = 1; i < row - 1; i++) {
+    for (let j = 1; j < col - 1; j++) {
+      dfs(i, j, true);
+    }
+  }
+
+  function dfs(i, j, isCounting) {
+    let condition = i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] === 0;
+
+    if (condition) return;
+    if (isCounting) count++;
+
+    grid[i][j] = 0;
+
+    dfs(i - 1, j, isCounting);
+    dfs(i + 1, j, isCounting);
+    dfs(i, j - 1, isCounting);
+    dfs(i, j + 1, isCounting);
+  }
+
+  return count;
+};
+```
+
 ### Rust
 
 dfs:
@@ -700,3 +757,4 @@ impl Solution {
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">
   <img src="../pics/网站星球宣传海报.jpg" width="1000"/>
 </a>
+

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/0047.参会dijkstra堆.md

@@ -655,6 +655,104 @@ int main() {
 
 ### Java 
 
+```Java 
+
+import java.util.*;
+
+class Edge {
+    int to;  // 邻接顶点
+    int val; // 边的权重
+
+    Edge(int to, int val) {
+        this.to = to;
+        this.val = val;
+    }
+}
+
+class MyComparison implements Comparator<Pair<Integer, Integer>> {
+    @Override
+    public int compare(Pair<Integer, Integer> lhs, Pair<Integer, Integer> rhs) {
+        return Integer.compare(lhs.second, rhs.second);
+    }
+}
+
+class Pair<U, V> {
+    public final U first;
+    public final V second;
+
+    public Pair(U first, V second) {
+        this.first = first;
+        this.second = second;
+    }
+}
+
+public class Main {
+    public static void main(String[] args) {
+        Scanner scanner = new Scanner(System.in);
+        int n = scanner.nextInt();
+        int m = scanner.nextInt();
+
+        List<List<Edge>> grid = new ArrayList<>(n + 1);
+        for (int i = 0; i <= n; i++) {
+            grid.add(new ArrayList<>());
+        }
+
+        for (int i = 0; i < m; i++) {
+            int p1 = scanner.nextInt();
+            int p2 = scanner.nextInt();
+            int val = scanner.nextInt();
+            grid.get(p1).add(new Edge(p2, val));
+        }
+
+        int start = 1;  // 起点
+        int end = n;    // 终点
+
+        // 存储从源点到每个节点的最短距离
+        int[] minDist = new int[n + 1];
+        Arrays.fill(minDist, Integer.MAX_VALUE);
+
+        // 记录顶点是否被访问过
+        boolean[] visited = new boolean[n + 1];
+
+        // 优先队列中存放 Pair<节点，源点到该节点的权值>
+        PriorityQueue<Pair<Integer, Integer>> pq = new PriorityQueue<>(new MyComparison());
+
+        // 初始化队列，源点到源点的距离为0，所以初始为0
+        pq.add(new Pair<>(start, 0));
+
+        minDist[start] = 0;  // 起始点到自身的距离为0
+
+        while (!pq.isEmpty()) {
+            // 1. 第一步，选源点到哪个节点近且该节点未被访问过（通过优先级队列来实现）
+            // <节点， 源点到该节点的距离>
+            Pair<Integer, Integer> cur = pq.poll();
+
+            if (visited[cur.first]) continue;
+
+            // 2. 第二步，该最近节点被标记访问过
+            visited[cur.first] = true;
+
+            // 3. 第三步，更新非访问节点到源点的距离（即更新minDist数组）
+            for (Edge edge : grid.get(cur.first)) { // 遍历 cur指向的节点，cur指向的节点为 edge
+                // cur指向的节点edge.to，这条边的权值为 edge.val
+                if (!visited[edge.to] && minDist[cur.first] + edge.val < minDist[edge.to]) { // 更新minDist
+                    minDist[edge.to] = minDist[cur.first] + edge.val;
+                    pq.add(new Pair<>(edge.to, minDist[edge.to]));
+                }
+            }
+        }
+
+        if (minDist[end] == Integer.MAX_VALUE) {
+            System.out.println(-1); // 不能到达终点
+        } else {
+            System.out.println(minDist[end]); // 到达终点最短路径
+        }
+    }
+}
+
+```
+
+
 ### Python
 
 ### Go