mirror of
https://github.com/youngyangyang04/leetcode-master.git
synced 2025-07-08 00:43:04 +08:00
Merge branch 'youngyangyang04:master' into master
This commit is contained in:
@ -183,6 +183,32 @@ class Solution {
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}
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}
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```
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```java
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// 解法2:通过判断path中是否存在数字,排除已经选择的数字
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class Solution {
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List<List<Integer>> result = new ArrayList<>();
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LinkedList<Integer> path = new LinkedList<>();
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public List<List<Integer>> permute(int[] nums) {
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if (nums.length == 0) return result;
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backtrack(nums, path);
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return result;
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}
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public void backtrack(int[] nums, LinkedList<Integer> path) {
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if (path.size() == nums.length) {
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result.add(new ArrayList<>(path));
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}
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for (int i =0; i < nums.length; i++) {
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// 如果path中已有,则跳过
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if (path.contains(nums[i])) {
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continue;
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}
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path.add(nums[i]);
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backtrack(nums, path);
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path.removeLast();
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}
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}
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}
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```
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Python:
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```python3
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@ -157,6 +157,28 @@ class Solution {
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}
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}
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```
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```java
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// 版本2
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class Solution {
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public int[][] merge(int[][] intervals) {
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LinkedList<int[]> res = new LinkedList<>();
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Arrays.sort(intervals, (o1, o2) -> Integer.compare(o1[0], o2[0]));
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res.add(intervals[0]);
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for (int i = 1; i < intervals.length; i++) {
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if (intervals[i][0] <= res.getLast()[1]) {
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int start = res.getLast()[0];
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int end = Math.max(intervals[i][1], res.getLast()[1]);
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res.removeLast();
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res.add(new int[]{start, end});
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}
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else {
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res.add(intervals[i]);
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}
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}
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return res.toArray(new int[res.size()][]);
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}
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}
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```
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Python:
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```python
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@ -200,6 +200,7 @@ public:
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Java:
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```java
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// 解法1
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class Solution {
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public int canCompleteCircuit(int[] gas, int[] cost) {
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int sum = 0;
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@ -221,7 +222,26 @@ class Solution {
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}
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}
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```
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```java
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// 解法2
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class Solution {
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public int canCompleteCircuit(int[] gas, int[] cost) {
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int curSum = 0;
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int totalSum = 0;
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int index = 0;
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for (int i = 0; i < gas.length; i++) {
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curSum += gas[i] - cost[i];
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totalSum += gas[i] - cost[i];
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if (curSum < 0) {
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index = (i + 1) % gas.length ;
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curSum = 0;
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}
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}
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if (totalSum < 0) return -1;
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return index;
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}
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}
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```
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Python:
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```python
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class Solution:
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@ -283,6 +303,35 @@ var canCompleteCircuit = function(gas, cost) {
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};
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```
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C:
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```c
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int canCompleteCircuit(int* gas, int gasSize, int* cost, int costSize){
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int curSum = 0;
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int i;
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int min = INT_MAX;
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//遍历整个数组。计算出每站的用油差。并将其与最小累加量比较
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for(i = 0; i < gasSize; i++) {
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int diff = gas[i] - cost[i];
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curSum += diff;
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if(curSum < min)
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min = curSum;
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}
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//若汽油总数为负数,代表无法跑完一环。返回-1
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if(curSum < 0)
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return -1;
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//若min大于等于0,说明每一天加油量比用油量多。因此从0出发即可
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if(min >= 0)
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return 0;
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//若累加最小值为负,则找到一个非零元素(加油量大于出油量)出发。返回坐标
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for(i = gasSize - 1; i >= 0; i--) {
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min+=(gas[i]-cost[i]);
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if(min >= 0)
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return i;
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}
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//逻辑上不会返回这个0
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return 0;
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}
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```
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-----------------------
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* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
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@ -7,7 +7,7 @@
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<p align="center"><strong>欢迎大家<a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!</strong></p>
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## 235. 二叉搜索树的最近公共祖先
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# 235. 二叉搜索树的最近公共祖先
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[力扣题目链接](https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-search-tree/)
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@ -21,14 +21,15 @@
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示例 1:
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输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
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输出: 6
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解释: 节点 2 和节点 8 的最近公共祖先是 6。
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* 输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
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* 输出: 6
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* 解释: 节点 2 和节点 8 的最近公共祖先是 6。
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示例 2:
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输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
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输出: 2
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解释: 节点 2 和节点 4 的最近公共祖先是 2, 因为根据定义最近公共祖先节点可以为节点本身。
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* 输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
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* 输出: 2
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* 解释: 节点 2 和节点 4 的最近公共祖先是 2, 因为根据定义最近公共祖先节点可以为节点本身。
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说明:
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@ -36,7 +37,9 @@
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* 所有节点的值都是唯一的。
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* p、q 为不同节点且均存在于给定的二叉搜索树中。
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## 思路
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# 思路
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做过[二叉树:公共祖先问题](https://programmercarl.com/0236.二叉树的最近公共祖先.html)题目的同学应该知道,利用回溯从底向上搜索,遇到一个节点的左子树里有p,右子树里有q,那么当前节点就是最近公共祖先。
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@ -58,6 +61,7 @@
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可以看出直接按照指定的方向,就可以找到节点4,为最近公共祖先,而且不需要遍历整棵树,找到结果直接返回!
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## 递归法
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递归三部曲如下:
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@ -111,7 +115,6 @@ if (cur->val > p->val && cur->val > q->val) {
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```
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if (递归函数(root->left)) return ;
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if (递归函数(root->right)) return ;
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```
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@ -128,7 +131,7 @@ left与right的逻辑处理;
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如果 cur->val 小于 p->val,同时 cur->val 小于 q->val,那么就应该向右遍历(目标区间在右子树)。
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```
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```CPP
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if (cur->val < p->val && cur->val < q->val) {
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TreeNode* right = traversal(cur->right, p, q);
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if (right != NULL) {
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@ -140,9 +143,9 @@ if (cur->val < p->val && cur->val < q->val) {
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剩下的情况,就是cur节点在区间(p->val <= cur->val && cur->val <= q->val)或者 (q->val <= cur->val && cur->val <= p->val)中,那么cur就是最近公共祖先了,直接返回cur。
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代码如下:
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```
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return cur;
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```
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那么整体递归代码如下:
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@ -216,7 +219,7 @@ public:
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灵魂拷问:是不是又被简单的迭代法感动到痛哭流涕?
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## 总结
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# 总结
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对于二叉搜索树的最近祖先问题,其实要比[普通二叉树公共祖先问题](https://programmercarl.com/0236.二叉树的最近公共祖先.html)简单的多。
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@ -225,10 +228,15 @@ public:
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最后给出了对应的迭代法,二叉搜索树的迭代法甚至比递归更容易理解,也是因为其有序性(自带方向性),按照目标区间找就行了。
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## 其他语言版本
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# 其他语言版本
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Java:
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## Java
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递归法:
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迭代法:
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```java
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class Solution {
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public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
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@ -246,15 +254,11 @@ class Solution {
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}
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```
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Python:
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```python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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## Python
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递归法:
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```python
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class Solution:
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def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
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if not root: return root //中
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@ -264,18 +268,14 @@ class Solution:
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return self.lowestCommonAncestor(root.right,p,q) //右
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else: return root
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```
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Go:
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> BSL法
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迭代法:
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## Go
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递归法:
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```go
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
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* }
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*/
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//利用BSL的性质(前序遍历有序)
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func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
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if root==nil{return nil}
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@ -287,34 +287,10 @@ func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
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}
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```
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> 普通法
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```go
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
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* }
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*/
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//递归会将值层层返回
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func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
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//终止条件
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if root==nil||root.Val==p.Val||root.Val==q.Val{return root}//最后为空或者找到一个值时,就返回这个值
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//后序遍历
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findLeft:=lowestCommonAncestor(root.Left,p,q)
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findRight:=lowestCommonAncestor(root.Right,p,q)
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//处理单层逻辑
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if findLeft!=nil&&findRight!=nil{return root}//说明在root节点的两边
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if findLeft==nil{//左边没找到,就说明在右边找到了
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return findRight
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}else {return findLeft}
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}
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```
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## JavaScript
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JavaScript版本:
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1. 使用递归的方法
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递归法:
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```javascript
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var lowestCommonAncestor = function(root, p, q) {
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// 使用递归的方法
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@ -336,7 +312,8 @@ var lowestCommonAncestor = function(root, p, q) {
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return root;
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};
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```
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2. 使用迭代的方法
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迭代法
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```javascript
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var lowestCommonAncestor = function(root, p, q) {
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// 使用迭代的方法
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@ -355,7 +332,6 @@ var lowestCommonAncestor = function(root, p, q) {
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```
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-----------------------
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* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
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* B站视频:[代码随想录](https://space.bilibili.com/525438321)
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|
@ -218,6 +218,30 @@ var findMinArrowShots = function(points) {
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};
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```
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C:
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```c
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int cmp(const void *a,const void *b)
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{
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return ((*((int**)a))[0] > (*((int**)b))[0]);
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}
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int findMinArrowShots(int** points, int pointsSize, int* pointsColSize){
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//将points数组作升序排序
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qsort(points, pointsSize, sizeof(points[0]),cmp);
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int arrowNum = 1;
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int i = 1;
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for(i = 1; i < pointsSize; i++) {
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//若前一个气球与当前气球不重叠,证明需要增加箭的数量
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if(points[i][0] > points[i-1][1])
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arrowNum++;
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else
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//若前一个气球与当前气球重叠,判断并更新最小的x_end
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points[i][1] = points[i][1] > points[i-1][1] ? points[i-1][1] : points[i][1];
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}
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return arrowNum;
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}
|
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```
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|
||||
-----------------------
|
||||
* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
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|
@ -110,15 +110,16 @@ class Solution {
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int len = nums.length;
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for (int i = 0; i < len; i++) {
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//从前向后遍历,遇到负数将其变为正数,同时K--
|
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if (nums[i] < 0 && k > 0) {
|
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if (nums[i] < 0 && K > 0) {
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nums[i] = -nums[i];
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k--;
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K--;
|
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}
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}
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// 如果K还大于0,那么反复转变数值最小的元素,将K用完
|
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if (k % 2 == 1) nums[len - 1] = -nums[len - 1];
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|
||||
|
||||
if (K % 2 == 1) nums[len - 1] = -nums[len - 1];
|
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return Arrays.stream(nums).sum();
|
||||
|
||||
}
|
||||
}
|
||||
```
|
||||
|
@ -137,7 +137,7 @@ dp[0][j] 和 dp[i][0] 都已经初始化了,那么其他下标应该初始化
|
||||
|
||||
```
|
||||
// 初始化 dp
|
||||
vector<vector<int>> dp(weight.size() + 1, vector<int>(bagWeight + 1, 0));
|
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vector<vector<int>> dp(weight.size(), vector<int>(bagWeight + 1, 0));
|
||||
for (int j = weight[0]; j <= bagWeight; j++) {
|
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dp[0][j] = value[0];
|
||||
}
|
||||
|
Reference in New Issue
Block a user