题目链接

动态规划

class Solution {public int uniquePathsWithObstacles(int[][] obstacleGrid) {int m = obstacleGrid.length;int n = obstacleGrid[0].length;int[][] dp = new int[m][n];// 遇到障碍则从(0,0)到达for(int i = 0; i < m && obstacleGrid[i][0] == 0; i++){dp[i][0] = 1;}for(int j = 0; j < n && obstacleGrid[0][j] == 0; j++){dp[0][j] = 1;}for(int i = 1; i < m; i++){for(int j = 1; j < n; j++){// 如果没有障碍则dp公式正确执行if(obstacleGrid[i][j] == 0){dp[i][j] = dp[i - 1][j] + dp[i][j - 1];}}}return dp[m - 1][n - 1];}
}

本题目是上一篇博客不同路径这道题目的升级版，整体动态规划思路相同，本题多了障碍这一条件约束。