【LeetCode】63. Unique Paths II（中等难度）

方法一 动态规划

由于 dp[i][j] = dp[i-1][j]+dp[i][j-1] 等于 dp[j]+=dp[j-1]，其中 dp[j] 就相当于 dp[i-1][j]，dp[i][j-1] 就相当于 dp[j-1]，就从二维压缩到一维了。

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

标签：Paths,obstacleGrid,int,res,II,++,length,63,dp
来源： https://blog.csdn.net/zeroheitao/article/details/120895243