Update

problems/0053.最大子序和.md

@@ -214,6 +214,7 @@ class Solution:
         return result
 
 ```
+贪心法
 ```python
 class Solution:
     def maxSubArray(self, nums):
@@ -226,8 +227,18 @@ class Solution:
             if count <= 0:  # 相当于重置最大子序起始位置，因为遇到负数一定是拉低总和
                 count = 0
         return result
-
-
+```
+动态规划
+```python
+class Solution:
+    def maxSubArray(self, nums: List[int]) -> int:
+        dp = [0] * len(nums)
+        dp[0] = nums[0]
+        res = nums[0]
+        for i in range(1, len(nums)):
+            dp[i] = max(dp[i-1] + nums[i], nums[i])
+            res = max(res, dp[i])
+        return res
 ```
 ### Go
 贪心法

problems/0070.爬楼梯完全背包版本.md

@@ -184,6 +184,16 @@ if __name__ == '__main__':
 
 ### Go：
 ```go
+package main
+
+import (
+	"bufio"
+	"fmt"
+	"os"
+	"strconv"
+	"strings"
+)
+
 func climbStairs(n int, m int) int {
     dp := make([]int, n+1)
     dp[0] = 1   

problems/0093.复原IP地址.md

@@ -467,9 +467,37 @@ class Solution:
         num = int(s[start:end+1])
         return 0 <= num <= 255
 
+回溯（版本三）
 
+```python
+class Solution:
+    def restoreIpAddresses(self, s: str) -> List[str]:
+        result = []
+        self.backtracking(s, 0, [], result)
+        return result
     
+    def backtracking(self, s, startIndex, path, result):
+        if startIndex == len(s):
+            result.append('.'.join(path[:]))
+            return
         
+        for i in range(startIndex, min(startIndex+3, len(s))):
+            # 如果 i 往后遍历了，并且当前地址的第一个元素是 0 ，就直接退出
+            if i > startIndex and s[startIndex] == '0':
+                break
+            # 比如 s 长度为 5，当前遍历到 i = 3 这个元素
+            # 因为还没有执行任何操作，所以此时剩下的元素数量就是 5 - 3 = 2 ，即包括当前的 i 本身
+            # path 里面是当前包含的子串，所以有几个元素就表示储存了几个地址
+            # 所以 (4 - len(path)) * 3 表示当前路径至多能存放的元素个数
+            # 4 - len(path) 表示至少要存放的元素个数
+            if (4 - len(path)) * 3 < len(s) - i or 4 - len(path) > len(s) - i:
+                break
+            if i - startIndex == 2:
+                if not int(s[startIndex:i+1]) <= 255:
+                    break
+            path.append(s[startIndex:i+1])
+            self.backtracking(s, i+1, path, result)
+            path.pop()
 ```
 
 ### Go 

problems/0110.平衡二叉树.md

@@ -609,10 +609,13 @@ class Solution:
         while stack:
             node = stack.pop()
             if node:
-                stack.append(node)
+                stack.append(node)              # 中
                 stack.append(None)
-                if node.left: stack.append(node.left)
-                if node.right: stack.append(node.right)
+                # 采用数组进行迭代，先将右节点加入，保证左节点能够先出栈
+                if node.right:                  # 右
+                    stack.append(node.right)
+                if node.left:                   # 左
+                    stack.append(node.left)
             else:
                 real_node = stack.pop()
                 left, right = height_map.get(real_node.left, 0), height_map.get(real_node.right, 0)

problems/0134.加油站.md

@@ -158,7 +158,7 @@ i从0开始累加rest[i]，和记为curSum，一旦curSum小于零，说明[0, i
 
 如果 curSum<0 说明 区间和1 + 区间和2 < 0， 那么 假设从上图中的位置开始计数curSum不会小于0的话，就是 区间和2>0。 
 
-区间和1 + 区间和2 < 0   同时 区间和2>0，只能说明区间和1 < 0， 那么就会从假设的箭头初就开始从新选择其实位置了。 
+区间和1 + 区间和2 < 0   同时 区间和2>0，只能说明区间和1 < 0， 那么就会从假设的箭头初就开始从新选择起始位置了。 
 
 
 **那么局部最优：当前累加rest[i]的和curSum一旦小于0，起始位置至少要是i+1，因为从i之前开始一定不行。全局最优：找到可以跑一圈的起始位置**。

problems/0151.翻转字符串里的单词.md

@@ -475,7 +475,45 @@ class Solution:
         words = words[::-1] # 反转单词
         return ' '.join(words) #列表转换成字符串
 ```
+(版本四) 将字符串转换为列表后，使用双指针去除空格
+```python
+class Solution:
+    def single_reverse(self, s, start: int, end: int):
+        while start < end:
+            s[start], s[end] = s[end], s[start]
+            start += 1
+            end -= 1
 
+    def reverseWords(self, s: str) -> str:
+        result = ""
+        fast = 0
+        # 1. 首先将原字符串反转并且除掉空格, 并且加入到新的字符串当中
+        # 由于Python字符串的不可变性，因此只能转换为列表进行处理
+        s = list(s)
+        s.reverse()
+        while fast < len(s):
+            if s[fast] != " ":
+                if len(result) != 0:
+                    result += " "
+                while s[fast] != " " and fast < len(s):
+                    result += s[fast]
+                    fast += 1
+            else:
+                fast += 1
+        # 2.其次将每个单词进行翻转操作
+        slow = 0
+        fast = 0
+        result = list(result)
+        while fast <= len(result):
+            if fast == len(result) or result[fast] == " ":
+                self.single_reverse(result, slow, fast - 1)
+                slow = fast + 1
+                fast += 1
+            else:
+                fast += 1
+
+        return "".join(result)
+```
 ### Go：
 
 版本一：

problems/0236.二叉树的最近公共祖先.md

@@ -454,7 +454,11 @@ impl Solution {
         p: Option<Rc<RefCell<TreeNode>>>,
         q: Option<Rc<RefCell<TreeNode>>>,
     ) -> Option<Rc<RefCell<TreeNode>>> {
-        if root == p || root == q || root.is_none() {
+        if root.is_none() {
+            return root;
+        }
+        if Rc::ptr_eq(root.as_ref().unwrap(), p.as_ref().unwrap()) 
+            || Rc::ptr_eq(root.as_ref().unwrap(), q.as_ref().unwrap()) {
             return root;
         }
         let left = Self::lowest_common_ancestor(

problems/0300.最长上升子序列.md

@@ -337,6 +337,29 @@ pub fn length_of_lis(nums: Vec<i32>) -> i32 {
 }
 ```
 
+### Cangjie:
+
+```cangjie
+func lengthOfLIS(nums: Array<Int64>): Int64 {
+    let n = nums.size
+    if (n <= 1) {
+        return n
+    }
+
+    let dp = Array(n, item: 1)
+    var res = 0
+    for (i in 1..n) {
+        for (j in 0..i) {
+            if (nums[i] > nums[j]) {
+                dp[i] = max(dp[i], dp[j] + 1)
+            }
+        }
+        res = max(dp[i], res)
+    }
+    return res
+}
+```
+
 
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">

problems/0343.整数拆分.md

@@ -243,6 +243,29 @@ class Solution {
     }
 }
 ```
+贪心
+```Java
+class Solution {
+    public int integerBreak(int n) {
+        // with 贪心
+        // 通过数学原理拆出更多的3乘积越大，则
+        /**
+         @Param: an int, the integer we need to break.
+         @Return: an int, the maximum integer after breaking
+         @Method: Using math principle to solve this problem
+         @Time complexity: O(1)
+        **/
+        if(n == 2) return 1;
+        if(n == 3) return 2;
+        int result = 1;
+        while(n > 4) {
+            n-=3;
+            result *=3;
+        }
+        return result*n;
+    }
+}
+```
 
 ### Python
 动态规划（版本一）

problems/0459.重复的子字符串.md

@@ -177,7 +177,7 @@ KMP算法中next数组为什么遇到字符不匹配的时候可以找到上一
 
 ![](https://code-thinking-1253855093.file.myqcloud.com/pics/20240913110841.png)
 
-这里有录友就想：如果字符串s 是有是有最小重复子串p组成，最长相等前后缀就不能更长一些？ 例如这样： 
+这里有录友就想：如果字符串s 是由最小重复子串p组成，最长相等前后缀就不能更长一些？ 例如这样： 
 
 ![](https://code-thinking-1253855093.file.myqcloud.com/pics/20240913114348.png)
 
@@ -884,3 +884,4 @@ public int[] GetNext(string s)
 <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/0494.目标和.md

@@ -706,6 +706,57 @@ class Solution:
 ```
 
 ### Go
+二维dp
+```go
+func findTargetSumWays(nums []int, target int) int {
+    sum := 0
+    for _, v := range nums {
+        sum += v
+    }
+    if math.Abs(float64(target)) > float64(sum) {
+        return 0 // 此时没有方案
+    }
+    if (target + sum) % 2 == 1 {
+        return 0 // 此时没有方案
+    }
+    bagSize := (target + sum) / 2
+    
+    dp := make([][]int, len(nums))
+    for i := range dp {
+        dp[i] = make([]int, bagSize + 1)
+    }
+    
+    // 初始化最上行
+    if nums[0] <= bagSize {
+        dp[0][nums[0]] = 1
+    }
+    
+    // 初始化最左列，最左列其他数值在递推公式中就完成了赋值
+    dp[0][0] = 1
+    
+    var numZero float64
+    for i := range nums {
+        if nums[i] == 0 {
+            numZero++
+        }
+        dp[i][0] = int(math.Pow(2, numZero))
+    }
+    
+    // 以下遍历顺序行列可以颠倒
+    for i := 1; i < len(nums); i++ { // 行，遍历物品
+        for j := 0; j <= bagSize; j++ { // 列，遍历背包
+            if nums[i] > j {
+                dp[i][j] = dp[i-1][j]
+            } else {
+                dp[i][j] = dp[i-1][j] + dp[i-1][j-nums[i]]
+            }
+        }
+    }
+    return dp[len(nums)-1][bagSize]
+}
+```
+
+一维dp
 ```go
 func findTargetSumWays(nums []int, target int) int {
 	sum := 0

problems/0518.零钱兑换II.md

@@ -288,6 +288,7 @@ class Solution:
 
 ### Go：
 
+一维dp
 ```go
 func change(amount int, coins []int) int {
 	// 定义dp数组
@@ -306,6 +307,29 @@ func change(amount int, coins []int) int {
 	return dp[amount]
 }
 ```
+二维dp
+```go
+func change(amount int, coins []int) int {
+    dp := make([][]int, len(coins))
+    for i := range dp {
+        dp[i] = make([]int, amount + 1)
+        dp[i][0] = 1
+    }
+    for j := coins[0]; j <= amount; j++ {
+        dp[0][j] += dp[0][j-coins[0]]
+    }
+    for i := 1; i < len(coins); i++ {
+        for j := 1; j <= amount; j++ {
+            if j < coins[i] {
+                dp[i][j] = dp[i-1][j]
+            } else {
+                dp[i][j] = dp[i][j-coins[i]] + dp[i-1][j]
+            }
+        }
+    }
+    return dp[len(coins)-1][amount]
+}
+```
 
 ### Rust:
 

problems/0674.最长连续递增序列.md

@@ -492,6 +492,25 @@ int findLengthOfLCIS(int* nums, int numsSize) {
 }
 ```
 
+### Cangjie
+
+```cangjie
+func findLengthOfLCIS(nums: Array<Int64>): Int64 {
+    let n = nums.size
+    if (n <= 1) {
+        return n
+    }
+    let dp = Array(n, repeat: 1)
+    var res = 0
+    for (i in 1..n) {
+        if (nums[i] > nums[i - 1]) {
+            dp[i] = dp[i - 1] + 1
+        }
+        res = max(res, dp[i])
+    }
+    return res
+}
+```
 
 
 

problems/0718.最长重复子数组.md

@@ -581,6 +581,25 @@ int findLength(int* nums1, int nums1Size, int* nums2, int nums2Size) {
 }
 ```
 
+### Cangjie
+
+```cangjie
+func findLength(nums1: Array<Int64>, nums2: Array<Int64>): Int64 {
+    let n = nums1.size
+    let m = nums2.size
+    let dp = Array(n + 1, {_ => Array(m + 1, item: 0)})
+    var res = 0
+    for (i in 1..=n) {
+        for (j in 1..=m) {
+            if (nums1[i - 1] == nums2[j - 1]) {
+                dp[i][j] = dp[i - 1][j - 1] + 1
+            }
+            res = max(res, dp[i][j])
+        }
+    }
+    return res
+}
+```
 
 
 <p align="center">

problems/1143.最长公共子序列.md

@@ -399,6 +399,25 @@ int longestCommonSubsequence(char* text1, char* text2) {
 }
 ```
 
+### Cangjie
+
+```cangjie
+func longestCommonSubsequence(text1: String, text2: String): Int64 {
+    let n = text1.size
+    let m = text2.size
+    let dp = Array(n + 1, {_ => Array(m + 1, repeat: 0)})
+    for (i in 1..=n) {
+        for (j in 1..=m) {
+            if (text1[i - 1] == text2[j - 1]) {
+                dp[i][j] = dp[i - 1][j - 1] + 1
+            } else {
+                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
+            }
+        }
+    }
+    return dp[n][m]
+}
+```
 
 
 <p align="center">

problems/kamacoder/0053.寻宝-Kruskal.md

@@ -549,6 +549,62 @@ if __name__ == "__main__":
 
 ### Javascript
 
+```js
+function kruskal(v, edges) {
+    const father = Array.from({ length: v + 1 }, (_, i) => i)
+    
+    function find(u){
+        if (u === father[u]) {
+            return u
+        } else {
+            father[u] = find(father[u])
+            return father[u]
+        }
+        
+    }
+    
+    function isSame(u, v) {
+        let s = find(u)
+        let t = find(v)
+        return s === t
+    }
+    
+    function join(u, v) {
+        let s = find(u)
+        let t = find(v)
+        if (s !== t) {
+            father[s] = t
+        }
+    }
+    
+    edges.sort((a, b) => a[2] - b[2])
+    let result = 0
+    for (const [v1, v2, w] of edges) {
+        if (!isSame(v1, v2)) {
+            result += w
+            join(v1 ,v2)
+        } 
+    }
+    console.log(result)
+}
+
+
+async function main() {
+    const rl = require('readline').createInterface({ input: process.stdin })
+    const iter = rl[Symbol.asyncIterator]()
+    const readline = async () => (await iter.next()).value
+    const [v, e] = (await readline()).split(" ").map(Number)
+    const edges = []
+    for (let i = 0 ; i < e ; i++) {
+        edges.push((await readline()).split(" ").map(Number))
+    }
+    kruskal(v, edges)
+}
+
+
+main()
+```
+
 ### TypeScript
 
 ### PhP

problems/kamacoder/0053.寻宝-prim.md

@@ -693,6 +693,55 @@ if __name__ == "__main__":
 ### Rust
 
 ### Javascript
+```js
+function prim(v, edges) {
+    const grid = Array.from({ length: v + 1 }, () => new Array(v + 1).fill(10001)); // Fixed grid initialization
+    const minDist = new Array(v + 1).fill(10001)
+    const isInTree = new Array(v + 1).fill(false)
+    // 建構鄰接矩陣
+    for(const [v1, v2, w] of edges) {
+        grid[v1][v2] = w
+        grid[v2][v1] = w
+    }
+    // prim 演算法
+    for (let i = 1 ; i < v ; i++) {
+        let cur = -1
+        let tempMinDist = Number.MAX_VALUE
+        // 1. 尋找距離生成樹最近的節點
+        for (let j = 1 ; j < v + 1 ; j++) {
+            if (!isInTree[j] && minDist[j] < tempMinDist) {
+                tempMinDist = minDist[j]
+                cur = j
+            }
+        }
+        // 2. 將節點放入生成樹
+        isInTree[cur] = true
+        // 3. 更新非生成樹節點與生成樹的最短距離
+        for (let j = 1 ; j < v + 1 ; j++) {
+            if (!isInTree[j] && grid[cur][j] < minDist[j]) {
+                minDist[j] = grid[cur][j]
+            }
+        }
+    }
+    console.log(minDist.slice(2).reduce((acc, cur) => acc + cur, 0))
+}
+
+
+async function main() {
+    const rl = require('readline').createInterface({ input: process.stdin })
+    const iter = rl[Symbol.asyncIterator]()
+    const readline = async () => (await iter.next()).value
+    const [v, e] = (await readline()).split(" ").map(Number)
+    const edges = []
+    for (let i = 0 ; i < e ; i++) {
+        edges.push((await readline()).split(" ").map(Number))
+    }
+    prim(v, edges)
+}
+
+
+main()
+```
 
 ### TypeScript
 

problems/kamacoder/0100.岛屿的最大面积.md

@@ -480,7 +480,84 @@ const bfs = (graph, visited, x, y) => {
 })()
 ```
 
+```javascript
 
+// 深搜版
+
+const r1 = require('readline').createInterface({ input: process.stdin });
+// 创建readline接口
+let iter = r1[Symbol.asyncIterator]();
+// 创建异步迭代器
+const readline = async () => (await iter.next()).value;
+
+let graph // 地图
+let N, M // 地图大小
+let visited // 访问过的节点
+let result = 0 // 最大岛屿面积
+let count = 0 // 岛屿内节点数
+const dir = [[0, 1], [1, 0], [0, -1], [-1, 0]] //方向
+
+// 读取输入，初始化地图
+const initGraph = async () => {
+  let line = await readline();
+  [N, M] = line.split(' ').map(Number);
+  graph = new Array(N).fill(0).map(() => new Array(M).fill(0))
+  visited = new Array(N).fill(false).map(() => new Array(M).fill(false))
+
+  for (let i = 0; i < N; i++) {
+    line = await readline()
+    line = line.split(' ').map(Number)
+    for (let j = 0; j < M; j++) {
+      graph[i][j] = line[j]
+    }
+  }
+}
+
+/**
+ * @description: 从(x, y)开始深度优先遍历
+ * @param {*} graph 地图
+ * @param {*} visited 访问过的节点
+ * @param {*} x 开始搜索节点的下标
+ * @param {*} y 开始搜索节点的下标
+ * @return {*}
+ */
+const dfs = (graph, visited, x, y) => {
+  for (let i = 0; i < 4; i++) {
+    let nextx = x + dir[i][0]
+    let nexty = y + dir[i][1]
+    if(nextx < 0 || nextx >= N || nexty < 0 || nexty >= M) continue
+    if(!visited[nextx][nexty] && graph[nextx][nexty] === 1){
+      count++
+      visited[nextx][nexty] = true
+      dfs(graph, visited, nextx, nexty)
+    }
+  }
+}
+
+(async function () {
+
+  // 读取输入，初始化地图
+  await initGraph()
+
+  // 统计最大岛屿面积
+  for (let i = 0; i < N; i++) {
+    for (let j = 0; j < M; j++) {
+      if (!visited[i][j] && graph[i][j] === 1) { //遇到没有访问过的陆地
+        // 重新计算面积
+        count = 1
+        visited[i][j] = true
+
+        // 深度优先遍历，统计岛屿内节点数，并将岛屿标记为已访问
+        dfs(graph, visited, i, j)
+
+        // 更新最大岛屿面积
+        result = Math.max(result, count)
+      }
+    }
+  }
+  console.log(result);
+})()
+```
 
 ### TypeScript
 

problems/kamacoder/0101.孤岛的总面积.md

@@ -307,6 +307,71 @@ for i in range(n):
 
 print(count)
 ```
+
+```python
+direction = [[1, 0], [-1, 0], [0, 1], [0, -1]]
+result = 0
+
+# 深度搜尋
+def dfs(grid, y, x):
+    grid[y][x] = 0
+    global result 
+    result += 1
+    
+    for i, j in direction:
+        next_x = x + j
+        next_y = y + i
+        if (next_x < 0 or next_y < 0 or
+            next_x >= len(grid[0]) or next_y >= len(grid)
+        ):
+            continue
+        if grid[next_y][next_x] == 1 and not visited[next_y][next_x]:
+            visited[next_y][next_x] = True
+            dfs(grid, next_y, next_x)    
+
+
+# 讀取輸入值
+n, m = map(int, input().split())
+grid = []
+visited = [[False] * m for _ in range(n)]
+
+for i in range(n):
+    grid.append(list(map(int, input().split())))
+
+# 處理邊界
+for j in range(m):
+    # 上邊界
+    if grid[0][j] == 1 and not visited[0][j]: 
+        visited[0][j] = True
+        dfs(grid, 0, j)
+    # 下邊界
+    if grid[n - 1][j] == 1 and not visited[n - 1][j]: 
+        visited[n - 1][j] = True
+        dfs(grid, n - 1, j)
+    
+for i in range(n):
+    # 左邊界
+    if grid[i][0] == 1 and not visited[i][0]: 
+        visited[i][0] = True
+        dfs(grid, i, 0)
+    # 右邊界
+    if grid[i][m - 1] == 1 and not visited[i][m - 1]: 
+        visited[i][m - 1] = True
+        dfs(grid, i, m - 1)
+    
+# 計算孤島總面積
+result = 0  # 初始化，避免使用到處理邊界時所產生的累加值
+
+for i in range(n):
+    for j in range(m):
+        if grid[i][j] == 1 and not visited[i][j]:
+            visited[i][j] = True
+            dfs(grid, i, j)
+
+# 輸出孤島的總面積
+print(result)
+```
+
 ### Go
 
 ``` go

problems/kamacoder/0108.冗余连接.md

@@ -178,6 +178,45 @@ int main() {
 
 ### Python
 
+```python
+father = list()
+
+def find(u):
+    if u == father[u]:
+        return u
+    else:
+        father[u] = find(father[u])
+        return father[u]
+        
+def is_same(u, v):
+    u = find(u)
+    v = find(v)
+    return u == v
+    
+def join(u, v):
+    u = find(u)
+    v = find(v)
+    if u != v:
+        father[u] = v
+        
+if __name__ == "__main__":
+    # 輸入
+    n = int(input())
+    for i in range(n + 1):
+        father.append(i)
+    # 尋找冗余邊    
+    result = None
+    for i in range(n):
+        s, t = map(int, input().split())
+        if is_same(s, t):
+            result = str(s) + ' ' + str(t)
+        else:
+            join(s, t)
+        
+    # 輸出
+    print(result)
+```
+
 ### Go
 
 ### Rust

problems/kamacoder/0109.冗余连接II.md

@@ -250,9 +250,193 @@ int main() {
 ## 其他语言版本
 
 ### Java 
+```java
+import java.util.ArrayList;
+import java.util.List;
+import java.util.Scanner;
 
+public class Main {
+    static int n;
+    static int[] father = new int[1001]; // 并查集数组
+
+    // 并查集初始化
+    public static void init() {
+        for (int i = 1; i <= n; ++i) {
+            father[i] = i;
+        }
+    }
+
+    // 并查集里寻根的过程
+    public static int find(int u) {
+        if (u == father[u]) return u;
+        return father[u] = find(father[u]); // 路径压缩
+    }
+
+    // 将 v->u 这条边加入并查集
+    public static void join(int u, int v) {
+        u = find(u);
+        v = find(v);
+        if (u != v) {
+            father[v] = u; // 合并两棵树
+        }
+    }
+
+    // 判断 u 和 v 是否有同一个根
+    public static boolean same(int u, int v) {
+        return find(u) == find(v);
+    }
+
+    // 在有向图里找到删除的那条边，使其变成树
+    public static void getRemoveEdge(List<int[]> edges) {
+        init(); // 初始化并查集
+        for (int i = 0; i < n; i++) { // 遍历所有的边
+            if (same(edges.get(i)[0], edges.get(i)[1])) { // 如果构成有向环了，就是要删除的边
+                System.out.println(edges.get(i)[0] + " " + edges.get(i)[1]);
+                return;
+            } else {
+                join(edges.get(i)[0], edges.get(i)[1]);
+            }
+        }
+    }
+
+    // 删一条边之后判断是不是树
+    public static boolean isTreeAfterRemoveEdge(List<int[]> edges, int deleteEdge) {
+        init(); // 初始化并查集
+        for (int i = 0; i < n; i++) {
+            if (i == deleteEdge) continue;
+            if (same(edges.get(i)[0], edges.get(i)[1])) { // 如果构成有向环了，一定不是树
+                return false;
+            }
+            join(edges.get(i)[0], edges.get(i)[1]);
+        }
+        return true;
+    }
+
+    public static void main(String[] args) {
+        Scanner sc = new Scanner(System.in);
+        List<int[]> edges = new ArrayList<>(); // 存储所有的边
+
+        n = sc.nextInt(); // 顶点数
+        int[] inDegree = new int[n + 1]; // 记录每个节点的入度
+        for (int i = 0; i < n; i++) {
+            int s = sc.nextInt(); // 边的起点
+            int t = sc.nextInt(); // 边的终点
+            inDegree[t]++;
+            edges.add(new int[]{s, t}); // 将边加入列表
+        }
+
+        List<Integer> vec = new ArrayList<>(); // 记录入度为2的边（如果有的话就两条边）
+        // 找入度为2的节点所对应的边，注意要倒序，因为优先删除最后出现的一条边
+        for (int i = n - 1; i >= 0; i--) {
+            if (inDegree[edges.get(i)[1]] == 2) {
+                vec.add(i);
+            }
+        }
+
+        // 情况一、情况二
+        if (vec.size() > 0) {
+            // vec里的边已经按照倒叙放的，所以优先删 vec.get(0) 这条边
+            if (isTreeAfterRemoveEdge(edges, vec.get(0))) {
+                System.out.println(edges.get(vec.get(0))[0] + " " + edges.get(vec.get(0))[1]);
+            } else {
+                System.out.println(edges.get(vec.get(1))[0] + " " + edges.get(vec.get(1))[1]);
+            }
+            return;
+        }
+
+        // 处理情况三：明确没有入度为2的情况，一定有有向环，找到构成环的边返回即可
+        getRemoveEdge(edges);
+    }
+}
+```
 ### Python
 
+```python
+from collections import defaultdict
+
+father = list()
+
+
+def find(u):
+    if u == father[u]:
+        return u
+    else:
+        father[u] = find(father[u])
+        return father[u]
+        
+        
+def is_same(u, v):
+    u = find(u)
+    v = find(v)
+    return u == v
+    
+    
+def join(u, v):
+    u = find(u)
+    v = find(v)
+    if u != v:
+        father[u] = v
+    
+    
+def is_tree_after_remove_edge(edges, edge, n):
+    # 初始化并查集
+    global father 
+    father = [i for i in range(n + 1)]
+    
+    for i in range(len(edges)):
+        if i == edge:
+            continue
+        s, t = edges[i]
+        if is_same(s, t): # 成環，即不是有向樹
+            return False
+        else: # 將s,t放入集合中
+            join(s, t)
+    return True
+    
+
+def get_remove_edge(edges):
+    # 初始化并查集
+    global father
+    father = [i for i in range(n + 1)]
+    
+    for s, t in edges:
+        if is_same(s, t):
+            print(s, t)
+            return
+        else:
+            join(s, t)
+        
+
+if __name__ == "__main__":
+    # 輸入
+    n = int(input())
+    edges = list()
+    in_degree = defaultdict(int)
+    
+    for i in range(n):
+        s, t = map(int, input().split())
+        in_degree[t] += 1
+        edges.append([s, t])
+        
+    # 尋找入度為2的邊，並紀錄其下標(index)
+    vec = list()
+    for i in range(n - 1, -1, -1):
+        if in_degree[edges[i][1]] == 2:
+            vec.append(i)
+            
+    # 輸出
+    if len(vec) > 0:
+        # 情況一：刪除輸出順序靠後的邊 
+        if is_tree_after_remove_edge(edges, vec[0], n):
+            print(edges[vec[0]][0], edges[vec[0]][1])
+        # 情況二：只能刪除特定的邊
+        else:
+            print(edges[vec[1]][0], edges[vec[1]][1])
+    else:
+        # 情況三： 原圖有環
+        get_remove_edge(edges)
+```
+
 ### Go
 
 ### Rust