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Commit 7cfdb12

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Merge remote-tracking branch 'origin/master'
# Conflicts: # .idea/workspace.xml # README.md
2 parents 67c2adc + 522f0c7 commit 7cfdb12

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15 files changed

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‎.idea/workspace.xml‎

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‎README.md‎

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### 链表操作
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* [q2_两数相加](/src/链表操作/q2_两数相加)
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* [q19_删除链表的倒数第N个节点](/src/链表操作/q19_删除链表的倒数第N个节点)
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* [q61_旋转链表](/src/链表操作/q61_旋转链表)
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* [q138_复制带随机指针的链表](/src/链表操作/q138_复制带随机指针的链表)
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* [q206_反转链表](/src/链表操作/q206_反转链表)
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* [q54_螺旋矩阵](/src/数组操作/q54_螺旋矩阵)
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* [q73_矩阵置零](/src/数组操作/q73_矩阵置零)
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* [q945_使数组唯一的最小增量](/src/数组操作/q945_使数组唯一的最小增量)
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### 栈相关
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* [q20_有效的括号](/src/栈相关/q20_有效的括号)
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* [q32_最长有效括号](/src/栈相关/q32_最长有效括号)
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* [q155_最小栈](/src/栈相关/q155_最小栈)
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* [q224_基本计算器](/src/栈相关/q224_基本计算器)
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* [q316_去除重复字母](/src/栈相关/q316_去除重复字母)
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* [q21_合并两个有序链表](/src/递归/q21_合并两个有序链表)
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* [q101_对称二叉树](/src/递归/q101_对称二叉树)
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* [q104_二叉树的最大深度](/src/递归/q104_二叉树的最大深度)
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* [q226_翻转二叉树](/src/递归/q226_翻转二叉树)
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* [q236_二叉树的最近公共祖先](/src/递归/q236_二叉树的最近公共祖先)
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### 分治法/二分法
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* [q23_合并K个排序链表](/src/分治法/q23_合并K个排序链表)
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* [q33_搜索旋转排序数组](/src/分治法/q33_搜索旋转排序数组)
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* [q34_在排序数组中查找元素的第一个和最后一个位置](/src/分治法/q34_在排序数组中查找元素的第一个和最后一个位置)
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### 动态规划
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* [q5_最长回文子串](/src/动态规划/q5_最长回文子串)
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* [q53_最大子序和](/src/动态规划/q53_最大子序和)
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* [q64_最小路径和](/src/动态规划/q64_最小路径和)
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* [q70_爬楼梯](/src/动态规划/q70_爬楼梯)
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* [q118_杨辉三角](/src/动态规划/q118_杨辉三角)
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* [q300_最长上升子序列](/src/动态规划/q300_最长上升子序列)
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* [q746_使用最小花费爬楼梯](/src/动态规划/q746_使用最小花费爬楼梯)

‎Rocket.md‎

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因为当Server端收到Client端的SYN连接请求报文后,可以直接发送SYN+ACK报文。其中ACK报文是用来应答的,SYN报文是用来同步的。但是关闭连接时,当Server端收到FIN报文时,很可能并不会立即关闭SOCKET,所以只能先回复一个ACK报文,告诉Client端,"你发的FIN报文我收到了"。只有等到我Server端所有的报文都发送完了,我才能发送FIN报文,因此不能一起发送。故需要四步握手。
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## 数据结构与算法
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### 排序算法
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package 分治法.q34_在排序数组中查找元素的第一个和最后一个位置;
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/**
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* 二分法 o(log(n))
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*/
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public class Solution {
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public int[] searchRange(int[] nums, int target) {
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if (nums == null || nums.length < 1) {
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return new int[]{-1, -1};
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}
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int midIndex = find(0, nums.length - 1, nums, target);
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int[] rs = new int[2];
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rs[0] = midIndex;
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rs[1] = midIndex;
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if (midIndex == -1) {
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return rs;
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}
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while (nums[rs[0]] == target && rs[0] > 0) {
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int temp = find(0, rs[0] - 1, nums, target);
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if (temp == -1) {
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break;
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} else {
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rs[0] = temp;
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}
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}
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while (nums[rs[1]] == target && rs[1] < nums.length - 1) {
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int temp = find(rs[1] + 1, nums.length - 1, nums, target);
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if (temp == -1) {
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break;
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} else {
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rs[1] = temp;
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}
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}
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return rs;
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}
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public int find(int beginIndex, int endIndex, int[] nums, int target) {
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if (beginIndex == endIndex) {
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if (nums[beginIndex] == target) {
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return beginIndex;
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} else {
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return -1;
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}
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}
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int mid = (endIndex - beginIndex) / 2 + beginIndex;
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if (nums[mid] > target) {
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return find(beginIndex, mid, nums, target);
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} else if (nums[mid] < target) {
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return find(mid + 1, endIndex, nums, target);
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} else {
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return mid;
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}
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}
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public static void main(String[] args) {
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new Solution().searchRange(new int[]{2, 2}, 2);
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}
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}
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package 动态规划.q64_最小路径和;
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/**
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* 动态规划 dp(j)=grid(i,j)+min(dp(j),dp(j+1)) o(m*n)
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*/
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public class Solution {
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public int minPathSum(int[][] grid) {
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int[] dp = new int[grid[0].length];
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for (int i = grid.length - 1; i >= 0; i--) {
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for (int j = grid[0].length - 1; j >= 0; j--) {
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if (i == grid.length - 1 && j != grid[0].length - 1) {
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dp[j] = grid[i][j] + dp[j + 1];
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} else if (j == grid[0].length - 1 && i != grid.length - 1) {
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dp[j] = grid[i][j] + dp[j];
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} else if (j != grid[0].length - 1 && i != grid.length - 1) {
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dp[j] = grid[i][j] + Math.min(dp[j], dp[j + 1]);
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} else {
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dp[j] = grid[i][j];
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}
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}
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}
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return dp[0];
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}
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}
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package 动态规划.q70_爬楼梯;
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/**
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* 动态规划 dp[i]表示到达第i阶的方法总数dp[i]=dp[i−1]+dp[i−2] o(n)
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*/
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public class Solution {
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public int climbStairs(int n) {
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if (n == 1) {
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return 1;
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}
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int[] dp = new int[n + 1];
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dp[1] = 1;
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dp[2] = 2;
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for (int i = 3; i <= n; i++) {
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dp[i] = dp[i - 1] + dp[i - 2];
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}
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return dp[n];
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}
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}
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package 数组操作.q945_使数组唯一的最小增量;
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import java.util.Arrays;
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/**
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* 先排序再遍历一次 o(n*log(n))
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*/
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public class Solution {
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public int minIncrementForUnique(int[] A) {
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if (A == null || A.length == 0 || A.length == 1) {
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return 0;
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}
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int rs = 0;
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Arrays.sort(A);
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int t = A[0];
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for (int i = 1; i < A.length; i++) {
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if (A[i] <= t) {
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rs = rs + t - A[i] + 1;
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A[i] = t + 1;
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}
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t = A[i];
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}
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return rs;
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}
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}
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package 栈相关.q155_最小栈;
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import java.util.Stack;
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/**
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* 不使用辅助栈,每次push两个元素
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*/
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public class MinStack {
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private Stack<Integer> stack;
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public MinStack() {
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stack = new Stack<>();
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}
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public void push(int x) {
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if (stack.isEmpty()) {
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stack.push(x);
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stack.push(x);
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} else {
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int tmp = stack.peek();
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stack.push(x);
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if (tmp < x) {
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stack.push(tmp);
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} else {
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stack.push(x);
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}
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}
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}
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public void pop() {
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stack.pop();
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stack.pop();
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}
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public int top() {
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return stack.get(stack.size() - 2);
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}
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public int getMin() {
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return stack.peek();
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}
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}
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package 栈相关.q32_最长有效括号;
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import java.util.Stack;
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/**
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* 利用索引栈 o(n)
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*/
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public class Solution {
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public int longestValidParentheses(String s) {
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if (s == null || s.length() < 2) {
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return 0;
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}
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int maxLen = 0;
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Stack<Integer> stack = new Stack<>();
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stack.push(-1);
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for (int i = 0; i < s.length(); i++) {
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char temp = s.charAt(i);
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if (temp == '(') {
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stack.push(i);
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} else {
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stack.pop();
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if (stack.empty()) {
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stack.push(i);
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} else {
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maxLen = Math.max(maxLen, i - stack.peek());
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}
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}
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}
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return maxLen;
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}
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public static void main(String[] args) {
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System.out.println(new Solution().longestValidParentheses(")()())"));
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}
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}
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package 递归.q104_二叉树的最大深度;
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/**
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* 递归 o(n)
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*/
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public class Solution {
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public int maxDepth(TreeNode root) {
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if (root == null) {
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return 0;
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} else {
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int leftHeight = maxDepth(root.left);
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int rightHeight = maxDepth(root.right);
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return Math.max(leftHeight, rightHeight) + 1;
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}
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}
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}

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