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https://github.com/youngyangyang04/leetcode-master.git
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Merge branch 'youngyangyang04:master' into master
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@ -250,6 +250,27 @@ var jump = function(nums) {
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};
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```
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### TypeScript
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```typescript
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function jump(nums: number[]): number {
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const length: number = nums.length;
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let curFarthestIndex: number = 0,
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nextFarthestIndex: number = 0;
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let curIndex: number = 0;
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let stepNum: number = 0;
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while (curIndex < length - 1) {
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nextFarthestIndex = Math.max(nextFarthestIndex, curIndex + nums[curIndex]);
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if (curIndex === curFarthestIndex) {
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curFarthestIndex = nextFarthestIndex;
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stepNum++;
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}
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curIndex++;
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}
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return stepNum;
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};
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```
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@ -154,6 +154,23 @@ var canJump = function(nums) {
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};
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```
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### TypeScript
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```typescript
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function canJump(nums: number[]): boolean {
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let farthestIndex: number = 0;
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let cur: number = 0;
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while (cur <= farthestIndex) {
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farthestIndex = Math.max(farthestIndex, cur + nums[cur]);
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if (farthestIndex >= nums.length - 1) return true;
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cur++;
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}
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return false;
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};
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```
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-----------------------
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
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@ -268,6 +268,18 @@ const maxProfit = (prices) => {
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};
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```
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TypeScript:
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```typescript
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function maxProfit(prices: number[]): number {
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let resProfit: number = 0;
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for (let i = 1, length = prices.length; i < length; i++) {
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resProfit += Math.max(prices[i] - prices[i - 1], 0);
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}
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return resProfit;
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};
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```
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C:
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```c
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@ -973,6 +973,10 @@ class MyLinkedList {
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// 处理头节点
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if (index === 0) {
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this.head = this.head!.next;
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// 如果链表中只有一个元素,删除头节点后,需要处理尾节点
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if (index === this.size - 1) {
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this.tail = null
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}
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this.size--;
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return;
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}
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@ -77,6 +77,61 @@ public:
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但这道题目的思路是很巧妙的,所以有必要介绍给大家做一做,感受一下。
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## 补充
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这里提供一种与[452.用最少数量的箭引爆气球](https://programmercarl.com/0452.用最少数量的箭引爆气球.html)、[435.无重叠区间](https://programmercarl.com/0435.无重叠区间.html)相同的思路。
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统计字符串中所有字符的起始和结束位置,记录这些区间(实际上也就是[435.无重叠区间](https://programmercarl.com/0435.无重叠区间.html)题目里的输入),**将区间按左边界从小到大排序,找到边界将区间划分成组,互不重叠。找到的边界就是答案。**
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```CPP
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class Solution {
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public:
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static bool cmp(vector<int> &a, vector<int> &b) {
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return a[0] < b[0];
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}
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// 记录每个字母出现的区间
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vector<vector<int>> countLabels(string s) {
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vector<vector<int>> hash(26, vector<int>(2, INT_MIN));
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vector<vector<int>> hash_filter;
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for (int i = 0; i < s.size(); ++i) {
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if (hash[s[i] - 'a'][0] == INT_MIN) {
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hash[s[i] - 'a'][0] = i;
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}
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hash[s[i] - 'a'][1] = i;
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}
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// 去除字符串中未出现的字母所占用区间
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for (int i = 0; i < hash.size(); ++i) {
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if (hash[i][0] != INT_MIN) {
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hash_filter.push_back(hash[i]);
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}
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}
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return hash_filter;
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}
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vector<int> partitionLabels(string s) {
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vector<int> res;
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// 这一步得到的 hash 即为无重叠区间题意中的输入样例格式:区间列表
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// 只不过现在我们要求的是区间分割点
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vector<vector<int>> hash = countLabels(s);
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// 按照左边界从小到大排序
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sort(hash.begin(), hash.end(), cmp);
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// 记录最大右边界
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int rightBoard = hash[0][1];
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int leftBoard = 0;
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for (int i = 1; i < hash.size(); ++i) {
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// 由于字符串一定能分割,因此,
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// 一旦下一区间左边界大于当前右边界,即可认为出现分割点
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if (hash[i][0] > rightBoard) {
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res.push_back(rightBoard - leftBoard + 1);
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leftBoard = hash[i][0];
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}
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rightBoard = max(rightBoard, hash[i][1]);
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}
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// 最右端
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res.push_back(rightBoard - leftBoard + 1);
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return res;
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}
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};
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```
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## 其他语言版本
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@ -208,6 +208,59 @@ int main() {
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## Java
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```Java
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public class Solution {
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// 节点类
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static class TreeNode {
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// 节点值
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int val;
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// 左节点
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TreeNode left;
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// 右节点
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TreeNode right;
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// 节点的构造函数(默认左右节点都为null)
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public TreeNode(int x) {
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this.val = x;
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this.left = null;
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this.right = null;
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}
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}
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/**
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* 根据数组构建二叉树
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* @param arr 树的数组表示
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* @return 构建成功后树的根节点
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*/
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public TreeNode constructBinaryTree(final int[] arr) {
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// 构建和原数组相同的树节点列表
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List<TreeNode> treeNodeList = arr.length > 0 ? new ArrayList<>(arr.length) : null;
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TreeNode root = null;
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// 把输入数值数组,先转化为二叉树节点列表
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for (int i = 0; i < arr.length; i++) {
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TreeNode node = null;
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if (arr[i] != -1) { // 用 -1 表示null
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node = new TreeNode(arr[i]);
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}
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treeNodeList.add(node);
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if (i == 0) {
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root = node;
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}
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}
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// 遍历一遍,根据规则左右孩子赋值就可以了
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// 注意这里 结束规则是 i * 2 + 2 < arr.length,避免空指针
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for (int i = 0; i * 2 + 2 < arr.length; i++) {
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TreeNode node = treeNodeList.get(i);
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if (node != null) {
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// 线性存储转连式存储关键逻辑
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node.left = treeNodeList.get(2 * i + 1);
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node.right = treeNodeList.get(2 * i + 2);
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}
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}
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return root;
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}
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}
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```
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