leetcode63—Unique Path II

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

Note: m and n will be at most 100.

Example 1:

Input:
[   [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:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right –> Right

想法：与Unique Path类似，只不过需要考虑障碍物，将有障碍物出现的地方，result设置为0.如何m-1,n-1位置有障碍物，直接返回0

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int row =  obstacleGrid.size();
        int col = obstacleGrid[0].size();
        int result[row][col];
        for(int i = 0 ; i < row ; i++){
            for(int j = 0 ; j < col ; j++){
                if(obstacleGrid[i][j] == 1){
                    result[i][j] = 0;
                }else{
                    if(i == 0 && j == 0){
                        result[i][j] = 1;
                    }else{
                        result[i][j] = 0;
                        if(i - 1 >= 0 ){
                            result[i][j] += result[i-1][j];
                        }
                        if(j - 1 >= 0 ){
                            result[i][j] += result[i][j-1];
                        }
                        }

                }

            }
        }
        return result[row-1][col-1];
    }
};

原文地址：https://www.cnblogs.com/tingweichen/p/10011326.html