Leetcode 63. Unique Paths II

1. Description

2. Solution

• Version 1

代码语言：javascript复制

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

• Version 2

代码语言：javascript复制

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

version

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