1 63 Unique Paths II

1.1 题目描述

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 the diagram below).

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and space is marked as 1 and 0 respectively in the grid.

输入：整数数组grid,表示一个mxn的方格,grid[i][j]=1表示是障碍，不能通过；grid[i][j]=0表示可以通过。
输出：能够从左上角走到右下角的不重复的路径数
规则：只能向下或者向右走

Input: obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
Output: 2
Explanation: There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:

Right -> Right -> Down -> Down
Down -> Down -> Right -> Right

1.2 动态规划解决

class Solution {public int uniquePathsWithObstacles(int[][] obstacleGrid) {if(obstacleGrid[0][0]==1) return 0;int m = obstacleGrid.length;int n = obstacleGrid[0].length;int[][] dp = new int[m][n];dp[0][0] = (obstacleGrid[0][0]==1?0:1);for(int j=1;j<n;j++){dp[0][j]= (obstacleGrid[0][j]==1?0:dp[0][j-1]);}for(int i=1;i<m;i++){dp[i][0] = (obstacleGrid[i][0]==1?0:dp[i-1][0]);}for(int i=1;i<m;i++){for(int j=1;j<n;j++){dp[i][j] = (obstacleGrid[i][j]==0?dp[i-1][j]+dp[i][j-1]:0);}}return dp[m-1][n-1];}
}

2 64. Minimum Path Sum

2.1 题目理解

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.
输入：整数数组grid,表示一个mxn的方格,grid[i][j]表示通过方格的代价。
输出：能够从左上角走到右下角的最小代价
规则：只能向下或者向右走
Input: grid = [[1,3,1],[1,5,1],[4,2,1]]
Output: 7
Explanation: Because the path 1 → 3 → 1 → 1 → 1 minimizes the sum.

2.2 动态规划

class Solution {public int minPathSum(int[][] grid) {int m = grid.length;int n = grid[0].length;int[][] dp = new int[m][n];dp[0][0] = grid[0][0];for(int i=1;i<m;i++){dp[i][0] = dp[i-1][0]+grid[i][0];}for(int j=1;j<n;j++){dp[0][j] = dp[0][j-1]+grid[0][j];}for(int i=1;i<m;i++){for(int j=1;j<n;j++){dp[i][j] = Math.min(dp[i-1][j],dp[i][j-1])+grid[i][j];}}return dp[m-1][n-1];}
}

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