1. 原题链接 https://leetcode.com/problems/unique-paths-ii/description/…

题目: 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'…

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…

63. 不同路径 II 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 来表示. 说明:m 和 n 的值均不超过 100. 示例 1: 输入: [ [0,0,0], [0,1,0], [0,0,0] ] 输出: 2…

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…

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…

原题 思路: 用到dp的思想,到row,col点路径数量 : path[row][col]=path[row][col-1]+path[row-1][col]; 遍历row*col,如果map[row][col]为1,则将其置为0:如果非1,则进行上述公式. 最后返回path[终点row][终点col]的值即为解. 一开始的代码,beat 44%,效率不高 class Solution { public: int uniquePathsWithObstacles(vector<vector<i…

1. 原题链接 https://leetcode.com/problems/spiral-matrix-ii/description/ 2. 题目要求 给定一个正整数n,求出从1到n平方的螺旋矩阵.例如n为3,构成的螺旋矩阵如下图所示 3. 解题思路 该题与54题Spiral Matrix的解题思路大致相同,同样是一个while循环内,确定螺旋矩阵一个圈上的所有元素. 采用一个计数器count,count初始为1,每确定一个位置上的元素,count加1, 4. 代码实现 public class…

Given an array of integers, every element appears three times except for one. Find that single one. Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 自己不会做... 搜答案,发现思路真是太巧妙了 关键:统计每一位上出现1的次…

62. Unique Paths class Solution { public: int uniquePaths(int m, int n) { || n <= ) ; vector<vector<int> > dp(m,vector<int>(n)); dp[][] = ; ;i < m;i++) dp[i][] = ; ;i < n;i++) dp[][i] = ; ;i < m;i++){ ;j < n;j++){ dp[i][j]…