From 94bdecfc2eeab67d6657fb6c688478638a5b86fc Mon Sep 17 00:00:00 2001
From: dam <1782067308@qq.com>
Date: Fri, 29 Nov 2024 15:56:34 +0800
Subject: [PATCH 1/8] =?UTF-8?q?=E6=B7=BB=E5=8A=A0474.=20=E4=B8=80=E5=92=8C?=
=?UTF-8?q?=E9=9B=B6=20java=20=E4=B8=89=E7=BB=B4DP=E6=95=B0=E7=BB=84?=
=?UTF-8?q?=E5=AE=9E=E7=8E=B0=E4=BB=A3=E7=A0=81?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0474.一和零.md | 65 +++++++++++++++++++++++++++++++++++++-
1 file changed, 64 insertions(+), 1 deletion(-)
diff --git a/problems/0474.一和零.md b/problems/0474.一和零.md
index e514e729..9d24f014 100644
--- a/problems/0474.一和零.md
+++ b/problems/0474.一和零.md
@@ -261,8 +261,70 @@ public:
## 其他语言版本
-
### Java
+
+三维DP数组实现
+
+```java
+class Solution {
+ public int findMaxForm(String[] strs, int m, int n) {
+ /// 数组有三个维度
+ // 第一个维度：取前面的几个字符串
+ // 第二个维度：0的数量限制（背包维度 1 容量）
+ // 第三个维度：1的数量限制（背包维度 2 容量）
+ int[][][] dpArr = new int[strs.length][m + 1][n + 1];
+
+ /// 初始化dpArr数组
+ // 计算第一个字符串的零数量和1数量
+ int zeroNum = 0;
+ int oneNum = 0;
+ for (char c : strs[0].toCharArray()) {
+ if (c == '0') {
+ zeroNum++;
+ } else {
+ oneNum++;
+ }
+ }
+ // 当0数量、1数量都容得下第一个字符串时，将DP数组的相应位置初始化为1，因为当前的子集数量为1
+ for (int j = zeroNum; j <= m; j++) {
+ for (int k = oneNum; k <= n; k++) {
+ dpArr[0][j][k] = 1;
+ }
+ }
+ /// 依次填充加入第i个字符串之后的DP数组
+ for (int i = 1; i < strs.length; i++) {
+ zeroNum = 0;
+ oneNum = 0;
+ for (char c : strs[i].toCharArray()) {
+ if (c == '0') {
+ zeroNum++;
+ } else {
+ oneNum++;
+ }
+ }
+ for (int j = 0; j <= m; j++) {
+ for (int k = 0; k <= n; k++) {
+ if (j >= zeroNum && k >= oneNum) {
+ // --if-- 当0数量维度和1数量维度的容量都大于等于当前字符串的0数量和1数量时，才考虑是否将当前字符串放入背包
+ // 不放入第i个字符串，子集数量仍为 dpArr[i - 1][j][k]
+ // 放入第i个字符串，需要在0维度腾出 zeroNum 个容量，1维度腾出 oneNum 个容量，然后放入当前字符串，即 dpArr[i - 1][j - zeroNum][k - oneNum] + 1)
+ dpArr[i][j][k] = Math.max(dpArr[i - 1][j][k], dpArr[i - 1][j - zeroNum][k - oneNum] + 1);
+ } else {
+ // --if-- 无法放入第i个字符串，子集数量仍为 dpArr[i - 1][j][k]
+ dpArr[i][j][k] = dpArr[i - 1][j][k];
+ }
+ }
+ }
+ }
+ return dpArr[dpArr.length - 1][m][n];
+ }
+}
+```
+
+
+
+二维DP数组实现
+
```Java
class Solution {
public int findMaxForm(String[] strs, int m, int n) {
@@ -682,3 +744,4 @@ public class Solution
+
From 1d0333f59634bc68b9c4f253f95a2ef8f7a5009e Mon Sep 17 00:00:00 2001
From: Anqi Li <103280095+iqna126@users.noreply.github.com>
Date: Wed, 15 Jan 2025 17:16:40 -0800
Subject: [PATCH 2/8] =?UTF-8?q?=E7=BB=990054.=E6=9B=BF=E6=8D=A2=E6=95=B0?=
=?UTF-8?q?=E5=AD=97.md=20=E5=8A=A0=E5=85=A5python=E8=A7=A3=E6=B3=95?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/kamacoder/0054.替换数字.md | 40 +++++++++++++++++++++++++
1 file changed, 40 insertions(+)
diff --git a/problems/kamacoder/0054.替换数字.md b/problems/kamacoder/0054.替换数字.md
index de0ab1a3..e4b5c43f 100644
--- a/problems/kamacoder/0054.替换数字.md
+++ b/problems/kamacoder/0054.替换数字.md
@@ -215,6 +215,46 @@ public class Main {
}
```
+### Python：
+```python
+class Solution(object):
+ def subsitute_numbers(self, s):
+ """
+ :type s: str
+ :rtype: str
+ """
+
+ count = sum(1 for char in s if char.isdigit()) # 统计数字的个数
+ expand_len = len(s) + (count * 5) # 计算扩充后字符串的大小， x->number， 每有一个数字就要增加五个长度
+ res = [''] * expand_len
+
+ new_index = expand_len - 1 # 指向扩充后字符串末尾
+ old_index = len(s) - 1 # 指向原字符串末尾
+
+ while old_index >= 0: # 从后往前， 遇到数字替换成“number”
+ if s[old_index].isdigit():
+ res[new_index-5:new_index+1] = "number"
+ new_index -= 6
+ else:
+ res[new_index] = s[old_index]
+ new_index -= 1
+ old_index -= 1
+
+ return "".join(res)
+
+if __name__ == "__main__":
+ solution = Solution()
+
+ while True:
+ try:
+ s = input()
+ result = solution.subsitute_numbers(s)
+ print(result)
+ except EOFError:
+ break
+
+```
+
### Go：
````go
package main
From da742feaa3e324c9489bedaf867004cae583cc90 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E2=80=98windscape=E2=80=99?= <2462269317@qq.com>
Date: Thu, 23 Jan 2025 11:47:27 +0800
Subject: [PATCH 3/8] =?UTF-8?q?=E4=BF=AE=E6=94=B90112.=E8=B7=AF=E5=BE=84?=
=?UTF-8?q?=E6=80=BB=E5=92=8C.md=E7=9A=84Java=E7=89=88=E6=9C=AC=E4=BB=A3?=
=?UTF-8?q?=E7=A0=81=E7=9A=84=E5=AD=97=E6=AF=8D=E5=B0=8F=E5=86=99=E9=97=AE?=
=?UTF-8?q?=E9=A2=98?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0112.路径总和.md | 59 ++++++++++++++++++-----------------
1 file changed, 31 insertions(+), 28 deletions(-)
diff --git a/problems/0112.路径总和.md b/problems/0112.路径总和.md
index b97013e6..141967f5 100644
--- a/problems/0112.路径总和.md
+++ b/problems/0112.路径总和.md
@@ -309,25 +309,25 @@ public:
0112.路径总和
```java
-class solution {
- public boolean haspathsum(treenode root, int targetsum) {
+class Solution {
+ public boolean hasPathSum(TreeNode root, int targetSum) {
if (root == null) {
return false;
}
- targetsum -= root.val;
+ targetSum -= root.val;
// 叶子结点
if (root.left == null && root.right == null) {
- return targetsum == 0;
+ return targetSum == 0;
}
if (root.left != null) {
- boolean left = haspathsum(root.left, targetsum);
- if (left) { // 已经找到
+ boolean left = hasPathSum(root.left, targetSum);
+ if (left) { // 已经找到，提前返回
return true;
}
}
if (root.right != null) {
- boolean right = haspathsum(root.right, targetsum);
- if (right) { // 已经找到
+ boolean right = hasPathSum(root.right, targetSum);
+ if (right) { // 已经找到，提前返回
return true;
}
}
@@ -336,16 +336,16 @@ class solution {
}
// lc112 简洁方法
-class solution {
- public boolean haspathsum(treenode root, int targetsum) {
+class Solution {
+ public boolean hasPathSum(TreeNode root, int targetSum) {
if (root == null) return false; // 为空退出
// 叶子节点判断是否符合
- if (root.left == null && root.right == null) return root.val == targetsum;
+ if (root.left == null && root.right == null) return root.val == targetSum;
// 求两侧分支的路径和
- return haspathsum(root.left, targetsum - root.val) || haspathsum(root.right, targetsum - root.val);
+ return hasPathSum(root.left, targetSum - root.val) || hasPathSum(root.right, targetSum - root.val);
}
}
```
@@ -353,22 +353,22 @@ class solution {
迭代
```java
-class solution {
- public boolean haspathsum(treenode root, int targetsum) {
+class Solution {
+ public boolean hasPathSum(TreeNode root, int targetSum) {
if(root == null) return false;
- stack stack1 = new stack<>();
- stack stack2 = new stack<>();
+ Stack stack1 = new Stack<>();
+ Stack stack2 = new Stack<>();
stack1.push(root);
stack2.push(root.val);
- while(!stack1.isempty()) {
+ while(!stack1.isEmpty()) {
int size = stack1.size();
for(int i = 0; i < size; i++) {
- treenode node = stack1.pop();
+ TreeNode node = stack1.pop();
int sum = stack2.pop();
// 如果该节点是叶子节点了，同时该节点的路径数值等于sum，那么就返回true
- if(node.left == null && node.right == null && sum == targetsum) {
+ if(node.left == null && node.right == null && sum == targetSum) {
return true;
}
// 右节点，压进去一个节点的时候，将该节点的路径数值也记录下来
@@ -387,8 +387,9 @@ class solution {
}
}
```
-```Java 統一迭代法
- public boolean hasPathSum(TreeNode root, int targetSum) {
+```Java
+class Solution {
+ public boolean hasPathSum(TreeNode root, int targetSum) {
Stack treeNodeStack = new Stack<>();
Stack sumStack = new Stack<>();
@@ -422,38 +423,39 @@ class solution {
}
return false;
}
+}
```
0113.路径总和-ii
```java
-class solution {
- public List> pathsum(TreeNode root, int targetsum) {
+class Solution {
+ public List> pathSum(TreeNode root, int targetSum) {
List> res = new ArrayList<>();
if (root == null) return res; // 非空判断
List path = new LinkedList<>();
- preorderdfs(root, targetsum, res, path);
+ preOrderDfs(root, targetSum, res, path);
return res;
}
- public void preorderdfs(TreeNode root, int targetsum, List> res, List path) {
+ public void preOrderDfs(TreeNode root, int targetSum, List> res, List path) {
path.add(root.val);
// 遇到了叶子节点
if (root.left == null && root.right == null) {
// 找到了和为 targetsum 的路径
- if (targetsum - root.val == 0) {
+ if (targetSum - root.val == 0) {
res.add(new ArrayList<>(path));
}
return; // 如果和不为 targetsum，返回
}
if (root.left != null) {
- preorderdfs(root.left, targetsum - root.val, res, path);
+ preOrderDfs(root.left, targetSum - root.val, res, path);
path.remove(path.size() - 1); // 回溯
}
if (root.right != null) {
- preorderdfs(root.right, targetsum - root.val, res, path);
+ preOrderDfs(root.right, targetSum - root.val, res, path);
path.remove(path.size() - 1); // 回溯
}
}
@@ -1626,3 +1628,4 @@ public class Solution {
+
From 163c3f33d8d24735230b25ddb697d726f0d5c242 Mon Sep 17 00:00:00 2001
From: Anqi Li <103280095+iqna126@users.noreply.github.com>
Date: Thu, 23 Jan 2025 08:16:44 -0800
Subject: [PATCH 4/8] =?UTF-8?q?0054.=E6=9B=BF=E6=8D=A2=E6=95=B0=E5=AD=97.m?=
=?UTF-8?q?d=20=E5=9C=A8=E8=AE=B2=E8=A7=A3=E4=B8=AD=E5=8A=A0=E5=85=A5?=
=?UTF-8?q?=E4=BA=86python=20string=E4=B9=9F=E4=B8=8D=E5=8F=AF=E5=8F=98?=
=?UTF-8?q?=E7=9A=84=E8=A7=A3=E9=87=8A?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/kamacoder/0054.替换数字.md | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/problems/kamacoder/0054.替换数字.md b/problems/kamacoder/0054.替换数字.md
index e4b5c43f..f788d65b 100644
--- a/problems/kamacoder/0054.替换数字.md
+++ b/problems/kamacoder/0054.替换数字.md
@@ -21,7 +21,7 @@
## 思路
-如果想把这道题目做到极致，就不要只用额外的辅助空间了！ （不过使用Java刷题的录友，一定要使用辅助空间，因为Java里的string不能修改）
+如果想把这道题目做到极致，就不要只用额外的辅助空间了！ （不过使用Java和Python刷题的录友，一定要使用辅助空间，因为Java和Python里的string不能修改）
首先扩充数组到每个数字字符替换成 "number" 之后的大小。
From 27718a4dfca99b25ceda3ee2946130c29261be71 Mon Sep 17 00:00:00 2001
From: asnpro <920569392@qq.com>
Date: Fri, 24 Jan 2025 22:51:46 +0800
Subject: [PATCH 5/8] =?UTF-8?q?docs:=20=E4=B8=BA=200494.=E7=9B=AE=E6=A0=87?=
=?UTF-8?q?=E5=92=8C.md=20=E5=AE=8C=E5=96=84=20JavaDoc=20=E6=B3=A8?=
=?UTF-8?q?=E9=87=8A,=20=E6=B7=BB=E5=8A=A0=E5=8A=A8=E8=A7=84=E4=BA=94?=
=?UTF-8?q?=E9=83=A8=E6=9B=B2=E6=B3=A8=E9=87=8A,=20=E8=A7=84=E8=8C=83?=
=?UTF-8?q?=E9=83=A8=E5=88=86=E4=BB=A3=E7=A0=81=E6=A0=BC=E5=BC=8F,=20?=
=?UTF-8?q?=E4=BF=AE=E6=94=B9=E9=83=A8=E5=88=86=E5=8F=98=E9=87=8F=E5=90=8D?=
=?UTF-8?q?=E4=BB=A5=E5=A2=9E=E5=BC=BA=E4=BB=A3=E7=A0=81=E8=AF=AD=E4=B9=89?=
=?UTF-8?q?=E5=8C=96?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0494.目标和.md | 59 +++++++++++++++++++++++++++++++-------
1 file changed, 49 insertions(+), 10 deletions(-)
diff --git a/problems/0494.目标和.md b/problems/0494.目标和.md
index 4a1fc6ab..c38ba7e4 100644
--- a/problems/0494.目标和.md
+++ b/problems/0494.目标和.md
@@ -825,30 +825,69 @@ func abs(x int) int {
### JavaScript
```javascript
+/**
+ * 题目来源: {@link https://leetcode.cn/problems/target-sum/}
+ *
+ * 题解来源: {@link https://programmercarl.com/0494.%E7%9B%AE%E6%A0%87%E5%92%8C.html#%E7%AE%97%E6%B3%95%E5%85%AC%E5%BC%80%E8%AF%BE}
+ *
+ * 时间复杂度: O(n * C), C 为数组元素总和与目标值之和的一半
+ *
+ * 空间复杂度: O(C)
+ *
+ * @param { number[] } nums
+ * @param { number } target
+ * @return { number }
+ */
const findTargetSumWays = (nums, target) => {
+ // 原题目可转化为:
+ //
+ // 将所有元素划分为 2 个集合,
+ // 一个集合中包含所有要添加 "+" 号的元素, 一个集合中包含所有要添加 "-" 号的元素
+ //
+ // 设两个集合的元素和分别为 positive 和 negative, 所有元素总和为 sum, 那么有如下等式:
+ // positive + negative = sum (1)
+ // positive - negative = target (2)
+ // (1) 与 (2) 联立可得: positive = (sum + target) / 2,
+ // 所以如果能从原数组中取出若干个元素形成 1 个元素总和为 (sum + target) / 2 的集合,
+ // 就算得到了 1 种满足题意的组合方法
+ //
+ // 因此, 所求变为: 有多少种取法, 可使得容量为 (sum + target) / 2 的背包被装满?
- const sum = nums.reduce((a, b) => a+b);
+ const sum = nums.reduce((a, b) => a + b);
- if(Math.abs(target) > sum) {
+ if (Math.abs(target) > sum) {
return 0;
}
- if((target + sum) % 2) {
+ if ((target + sum) % 2) {
return 0;
}
- const halfSum = (target + sum) / 2;
-
- let dp = new Array(halfSum+1).fill(0);
+ const bagWeight = (target + sum) / 2;
+
+ // 1. dp 数组的含义
+ // dp[j]: 装满容量为 j 的背包, 有 dp[j] 种方法
+ let dp = new Array(bagWeight + 1).fill(0);
+
+ // 2. 递推公式
+ // dp[j] = Σ(dp[j - nums[j]]), (j ∈ [0, j] 且 j >= nums[j])
+ // 因为 dp[j - nums[j]] 表示: 装满容量为 j - nums[j] 背包有 dp[j - nums[j]] 种方法
+ // 而容量为 j - nums[j] 的背包只需要再将 nums[j] 放入背包就能使得背包容量达到 j
+ // 因此, 让背包容量达到 j 有 Σ(dp[j - nums[j]]) 种方法
+
+ // 3. dp 数组如何初始化
+ // dp[0] = 1, dp[1 ~ bagWeight] = 0
dp[0] = 1;
-
- for(let i = 0; i < nums.length; i++) {
- for(let j = halfSum; j >= nums[i]; j--) {
+
+ // 4. 遍历顺序
+ // 先物品后背包, 物品从前往后遍历, 背包容量从后往前遍历
+ for (let i = 0; i < nums.length; i++) {
+ for (let j = bagWeight; j >= nums[i]; j--) {
dp[j] += dp[j - nums[i]];
}
}
- return dp[halfSum];
+ return dp[bagWeight];
};
```
From e6698cbac457714427f4a9cba6ea2a3ea9eed94c Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E2=80=98windscape=E2=80=99?= <2462269317@qq.com>
Date: Sat, 25 Jan 2025 11:52:09 +0800
Subject: [PATCH 6/8] =?UTF-8?q?=E4=BF=AE=E6=94=B90530.=E4=BA=8C=E5=8F=89?=
=?UTF-8?q?=E6=90=9C=E7=B4=A2=E6=A0=91=E7=9A=84=E6=9C=80=E5=B0=8F=E7=BB=9D?=
=?UTF-8?q?=E5=AF=B9=E5=B7=AE.md=E7=9A=84Java=E7=89=88=E6=9C=AC=20?=
=?UTF-8?q?=E8=BF=9B=E8=A1=8C=E4=BA=86=E4=BB=A3=E7=A0=81=E6=A0=BC=E5=BC=8F?=
=?UTF-8?q?=E5=8C=96=E5=B9=B6=E6=B7=BB=E5=8A=A0=E4=BA=86=E7=BB=9F=E4=B8=80?=
=?UTF-8?q?=E8=BF=AD=E4=BB=A3=E6=B3=95=E7=9A=84=E6=B3=A8=E9=87=8A?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
.../0530.二叉搜索树的最小绝对差.md | 46 +++++++++++--------
1 file changed, 28 insertions(+), 18 deletions(-)
diff --git a/problems/0530.二叉搜索树的最小绝对差.md b/problems/0530.二叉搜索树的最小绝对差.md
index 2533a618..b6d08dbe 100644
--- a/problems/0530.二叉搜索树的最小绝对差.md
+++ b/problems/0530.二叉搜索树的最小绝对差.md
@@ -153,23 +153,27 @@ public:
递归
```java
class Solution {
- TreeNode pre;// 记录上一个遍历的结点
+ TreeNode pre; // 记录上一个遍历的结点
int result = Integer.MAX_VALUE;
+
public int getMinimumDifference(TreeNode root) {
- if(root==null)return 0;
- traversal(root);
- return result;
+ if (root == null)
+ return 0;
+ traversal(root);
+ return result;
}
- public void traversal(TreeNode root){
- if(root==null)return;
- //左
+
+ public void traversal(TreeNode root) {
+ if (root == null)
+ return;
+ // 左
traversal(root.left);
- //中
- if(pre!=null){
- result = Math.min(result,root.val-pre.val);
+ // 中
+ if (pre != null) {
+ result = Math.min(result, root.val - pre.val);
}
pre = root;
- //右
+ // 右
traversal(root.right);
}
}
@@ -182,22 +186,27 @@ class Solution {
TreeNode pre = null;
int result = Integer.MAX_VALUE;
- if(root != null)
+ if (root != null)
stack.add(root);
- while(!stack.isEmpty()){
+
+ // 中序遍历（左中右），由于栈先入后出，反序（右中左）
+ while (!stack.isEmpty()) {
TreeNode curr = stack.peek();
- if(curr != null){
+ if (curr != null) {
stack.pop();
- if(curr.right != null)
+ // 右
+ if (curr.right != null)
stack.add(curr.right);
+ // 中（先用null标记）
stack.add(curr);
stack.add(null);
- if(curr.left != null)
+ // 左
+ if (curr.left != null)
stack.add(curr.left);
- }else{
+ } else { // 中（遇到null再处理）
stack.pop();
TreeNode temp = stack.pop();
- if(pre != null)
+ if (pre != null)
result = Math.min(result, temp.val - pre.val);
pre = temp;
}
@@ -674,3 +683,4 @@ public class Solution
+
From c2e95f6c25b2c3f9ff61dc0891cc08360b676bb1 Mon Sep 17 00:00:00 2001
From: dam <1782067308@qq.com>
Date: Sun, 26 Jan 2025 12:08:15 +0800
Subject: [PATCH 7/8] =?UTF-8?q?108=E5=86=97=E4=BD=99=E8=BF=9E=E6=8E=A5=20j?=
=?UTF-8?q?ava=E5=AE=9E=E7=8E=B0?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/kamacoder/0108.冗余连接.md | 75 +++++++++++++++++++++++++
1 file changed, 75 insertions(+)
diff --git a/problems/kamacoder/0108.冗余连接.md b/problems/kamacoder/0108.冗余连接.md
index efbbb6d2..df3dd4de 100644
--- a/problems/kamacoder/0108.冗余连接.md
+++ b/problems/kamacoder/0108.冗余连接.md
@@ -176,6 +176,81 @@ int main() {
### Java
+```java
+import java.util.Scanner;
+
+public class Main {
+ private static int[] father;
+
+ public static void main(String[] args) {
+ Scanner scanner = new Scanner(System.in);
+ int pointNum = scanner.nextInt();
+ father = new int[pointNum + 1];
+ init();
+ for (int i = 0; i < pointNum; i++) {
+ join(scanner.nextInt(), scanner.nextInt());
+ }
+ }
+
+ /**
+ * 并查集初始化
+ */
+ private static void init() {
+ for (int i = 1; i < father.length; i++) {
+ // 让每个元素指向自己
+ father[i] = i;
+ }
+ }
+
+ /**
+ * 并查集寻根
+ *
+ * @param u
+ * @return
+ */
+ private static int find(int u) {
+ // 判断 u 是否等于自己，如果是的话，直接返回自己
+ // 如果不等于自己，就寻找根，寻找的时候，反复进行路径压缩
+ return u == father[u] ? u : (father[u] = find(father[u]));
+ }
+
+ /**
+ * 判断 u 和 v 是否同根
+ *
+ * @param u
+ * @param v
+ * @return
+ */
+ private static boolean isSame(int u, int v) {
+ return find(u) == find(v);
+ }
+
+ /**
+ * 添加 边 到并查集，v 指向 u
+ *
+ * @param u
+ * @param v
+ */
+ private static void join(int u, int v) {
+ // --if-- 如果两个点已经同根，说明他们的信息已经存储到并查集中了，直接返回即可
+ // 寻找u的根
+ int uRoot = find(u);
+ // 寻找v的根
+ int vRoot = find(v);
+ if (uRoot == vRoot) {
+ // --if-- 如果u,v的根相同，说明两者已经连接了，直接输出
+ System.out.println(u + " " + v);
+ return;
+ }
+ // --if-- 将信息添加到并查集
+ father[vRoot] = uRoot;
+ }
+
+}
+```
+
+
+
### Python
```python
From 333099a1268544306e303fcb779ffd801c0d0c92 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E2=80=98windscape=E2=80=99?= <2462269317@qq.com>
Date: Mon, 3 Feb 2025 20:13:41 +0800
Subject: [PATCH 8/8] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=8C=E5=8F=89?=
=?UTF-8?q?=E6=A0=91=E6=80=BB=E7=BB=93=E7=AF=87.md=E4=B8=80=E5=A4=84?=
=?UTF-8?q?=E5=88=97=E8=A1=A8=E7=BA=A7=E5=88=AB=E7=9A=84=E6=A0=BC=E5=BC=8F?=
=?UTF-8?q?=E9=97=AE=E9=A2=98?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/二叉树总结篇.md | 6 +++---
1 file changed, 3 insertions(+), 3 deletions(-)
diff --git a/problems/二叉树总结篇.md b/problems/二叉树总结篇.md
index 8db40d65..4794233a 100644
--- a/problems/二叉树总结篇.md
+++ b/problems/二叉树总结篇.md
@@ -92,10 +92,9 @@
* 递归：中序，双指针操作
* 迭代：模拟中序，逻辑相同
* [求二叉搜索树的众数](https://programmercarl.com/0501.二叉搜索树中的众数.html)
-
+
* 递归：中序，清空结果集的技巧，遍历一遍便可求众数集合
- * [二叉搜索树转成累加树](https://programmercarl.com/0538.把二叉搜索树转换为累加树.html)
-
+* [二叉搜索树转成累加树](https://programmercarl.com/0538.把二叉搜索树转换为累加树.html)
* 递归：中序，双指针操作累加
* 迭代：模拟中序，逻辑相同
@@ -163,3 +162,4 @@
+