解题思路

与LC62大致相似，多了一个路障,遇到路障时候跳过，若上i - 1或者j - 1是路障则不添加路线

首先dp[i][j]肯定是能到这个位置的所有路径

递推公式：dp[i][j] = dp[i - 1][j] + dp[i][j - 1]

初始化：dp[i][j] = 0

遍历顺序：顺序遍历

代码

Python

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class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m = len(obstacleGrid)
        n = len(obstacleGrid[0])
        if obstacleGrid[0][0] == 1: return 0
        dp = [[0] * n for _ in range(m)]
        dp[0][0] = 1
        for i in range(m):
            for j in range(n):
                if obstacleGrid[i][j] == 1: continue
                if i - 1 >= 0 and obstacleGrid[i - 1][j] != 1:
                    dp[i][j] += dp[i - 1][j]
                if j - 1 >= 0 and obstacleGrid[i][j - 1] != 1:
                    dp[i][j] += dp[i][j - 1]
        return dp[m - 1][n - 1]

C++

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

Go

func uniquePathsWithObstacles(obstacleGrid [][]int) int {
    m, n := len(obstacleGrid), len(obstacleGrid[0])
    dp := make([][]int, m)
    for i := range dp {
        dp[i] = make([]int, n)
    }
    if obstacleGrid[0][0] == 1 {return 0}
    dp[0][0] = 1
    for i := 0; i < m; i++ {
        for j := 0; j < n; j++ {
            if obstacleGrid[i][j] == 1 {continue}
            if i - 1 >= 0 && obstacleGrid[i - 1][j] != 1 {
                dp[i][j] += dp[i - 1][j]
            }
            if j - 1 >= 0 && obstacleGrid[i][j - 1] != 1 {
                dp[i][j] += dp[i][j - 1]
            }
        }
    }
    return dp[m - 1][n - 1]
}