引言 二维动态规划中最常见的是棋盘型二维动态规划. 即 func(i, j) 往往只和 func(i-1, j-1), func(i-1, j) 以及 func(i, j-1) 有关 这种情况下,时间复杂度 O(n*n),空间复杂度往往可以优化为O(n) 例题  1 Minimum Path Sum  Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right whi…

之所以将这三道题放在一起,是因为这三道题非常类似. 1. Minimum Path Sum 题目链接 题目要求: Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or righ…

97. Interleaving String Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2. For example, Given: s1 = "aabcc", s2 = "dbbca", When s3 = "aadbbcbcac", return true. When s3 = "aadbbbaccc", retu…

题意:现在要写m行代码,总共有n个文件,现在给出第i个文件每行会出现v[i]个bug,问你在bug少于b的条件下有多少种安排 分析:定义dp[i][j][k],i个文件,用了j行代码,有k个bug 状态转移为 1.在第i个文件,不写代码   dp[i][j][k]=dp[i-1][j][k] 2.在第i个文件,写代码      dp[i][j][k]+=dp[i][j-1][k-v[i]] 这题巧妙在于,既往i转移,又往j和k方向转移,这样我把它形容为二维动态规划 代码: #include <b…

唯一路径问题II Unique Paths II Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How many unique paths would there be? An obstacle and empty space is marked as 1 and 0 respectively in the grid. For example, There is…

Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 17880   Accepted: 6709 Description Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1…

Unique Paths https://oj.leetcode.com/problems/unique-paths/ A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to rea…

Leetcode之动态规划(DP)专题-63. 不同路径 II(Unique Paths II) 初级题目:Leetcode之动态规划(DP)专题-62. 不同路径(Unique Paths) 一个机器人位于一个 m x n 网格的左上角 (起始点在下图中标记为“Start” ). 机器人每次只能向下或者向右移动一步.机器人试图达到网格的右下角(在下图中标记为“Finish”). 现在考虑网格中有障碍物.那么从左上角到右下角将会有多少条不同的路径? 网格中的障碍物和空位置分别用 1 和 0 来表…

Leetcode之动态规划(DP)专题-64. 最小路径和(Minimum Path Sum) 给定一个包含非负整数的 m x n 网格,请找出一条从左上角到右下角的路径,使得路径上的数字总和为最小. 说明:每次只能向下或者向右移动一步. 示例: 输入: [   [1,3,1], [1,5,1], [4,2,1] ] 输出: 7 解释: 因为路径 1→3→1→1→1 的总和最小. 找从左上角0,0到右下角的最短路径. DP:我们每个点(x,y)都可以表示为dp[x][y] = max( grid…

303一维数组的升级版,方法就是用二维数组res存下从(0,0)到当前位置的sum,存的方法是动态规划,看着二维数组画圈比较好搞清楚其中的加减法 算子数组的sum的时候也是和存差不多的逻辑,就是某一部分加上另一部分,然后减去某一部分,逻辑画画圈就能看出来 比价重要的是动态规划存数的过程,以后二维数组问题应该会经常用 package com.DynamicProgramming; import java.util.HashMap; import java.util.Map; /** * Given…

题目描述与背景介绍 背景题目: [674. 最长连续递增序列]https://leetcode-cn.com/problems/longest-continuous-increasing-subsequence/ [300. 最长递增子序列]https://leetcode-cn.com/problems/longest-increasing-subsequence/ 这两个都是DP的经典题目,674比较简单. 代码: class Solution { public int findLength…

72. 编辑距离 再次验证leetcode的评判机有问题啊!同样的代码,第一次提交超时,第二次提交就通过了! 此题用动态规划解决. 这题一开始还真难到我了,琢磨半天没有思路.于是乎去了网上喵了下题解看到了动态规划4个字就赶紧回来了. 脑海中浮现了两个问题: 为什么能用动态规划呢?用动态规划怎么解? 先描述状态吧: f[i][j]表示word1中的[0,i] 与 word2中[0,j]的最少操作数. 实际上这时候就能看出来了,当一个状态计算完成时,即一个状态的操作方案(决策)确定时,不影响后面状态…

Stars Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/65536 K (Java/Others) Total Submission(s): 785    Accepted Submission(s): 335 Problem Description Yifenfei is a romantic guy and he likes to count the stars in the sky. To make the p…

Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How many unique paths would there be? An obstacle and empty space is marked as 1 and 0 respectively in the grid. For example, There is one obstacle in the middl…

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in t…

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example:Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1 return [ [5,4,11,2], [5,8,4,5] ] 这道二叉树路径之和在之前那道题 Path…

思路:a[i][j]表示j秒在i位置的数目,dp[i][j]表示j秒在i位置最大可以收到的数目. 转移方程:d[i][j]=max(dp[i-1][j],dp[i-1][j-1],dp[i-1][j+1]); #include<iostream> #include<string> #include<algorithm> #include<cstdlib> #include<cstdio> #include<set> #include&…

Palindrome Description A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a program which, given a string, determines the minimal number of characters to be inse…

问题描述:Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. 问题分析:参考路径问题,其实就是加权的路径问题,求最小的权值和.动态规划问题,核心递推公式,d[i][j] = min(d[i-1][j],d[i][j-1])+a[i][j]. publi…

/** * Given a m x n grid filled with non-negative numbers, * find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time. */ /* * 动态规划题目,思路是把每一个点的最…

description: Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: Example: Example: Input: [ [1,3,1], [1,5,1], [4,2,1] ] Output: 7 Explanation: Bec…

description: https://leetcode.com/problems/unique-paths/ 机器人从一堆方格的左上角走到右下角,只能往右或者往下走 ,问有几种走法,这个加了难度,在矩阵中加了障碍物 Note: Example: Example 1: Input: [ [0,0,0], [0,1,0], [0,0,0] ] Output: 2 Explanation: There is one obstacle in the middle of the 3x3 grid ab…

xiaoz 征婚,首先输入M,表示有M个操作. 借下来M行,对每一行   Ih a l     I 表示有一个MM报名,H是高度, a是活泼度,L是缘分. 或   Q h1 h2 a1 a2    求出身高在h1  h2  活泼度在a1  a2之间的最大缘分值. #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm>…

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time. Example 1: [[1,3,1], [1,5,1], [4,…

Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How many unique paths would there be? An obstacle and empty space is marked as 1 and 0 respectively in the grid. For example, There is one obstacle in the middl…

洛谷1387 dp题目,刚开始写的时候使用了前缀和加搜索,复杂度大概在O(n ^ 3)级别,感觉这么写还是比较对得起普及/提高-的难度的..后来看了题解区各位大神的题解,开始一脸mb,之后备受启发. 设dp[i][j]表示以(i, j)为右下点的正方形的最大边长,则转移方程如下: dp[i][j] = min{dp[i-1][j], dp[i][j-1], dp[i-1][j-1]} + 1 (a[i][j] == 1) dp[i][j] = 0 (a[i][j] == 0) 转移非常简单,但是…

在Path SUm 1中(http://www.cnblogs.com/hitkb/p/4242822.html) 我们采用栈的形式保存路径,每当找到符合的叶子节点,就将栈内元素输出.注意存在多条路径的情况. public List<List<Integer>> pathSum(TreeNode root, int sum) { List<List<Integer>>list=new ArrayList<>(); Stack<TreeNod…

Path Sum Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. For example: Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ \ 7 2 1 r…

[题目]D. Animals and Puzzle [题意]给定n*m的01矩阵,Q次询问某个子矩阵内的最大正方形全1子矩阵边长.n,m<=1000,Q<=10^6. [算法]动态规划DP+二维ST表 [题解]设f[i][j]为以(i,j)为右下角的最大正方形全1子矩阵. f[i][j]=min{ f[i-1][j-1] , f[i][j-1] , f[i-1][j] }+1 然后用二维ST表处理f[i][j]的子矩阵最小值. 对于每次询问,二分边长x,答案即子矩阵(x1+x-1,y1+x-1…

动态规划 题目分类 一维dp 矩阵型DP Unique Paths II : 矩阵型DP,求所有方法总数 Minimum Path Sum:矩阵型,求最大最小值 Triangle : 矩阵型,求最大最小值 Maximum Square :矩阵型,求最大最小值 Range Sum Query 2D - Immutable : 求和 Unique Paths II : 矩阵型DP,求所有方法总数 方法一:自顶向下 递归法,time limited class Solution { int helpe…