Description

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

分析

题目的意思是：一个机器人从左上角到右下角，问机器人能够到右下角的路径的数目。

• dp[i][j]代表dp[0][0]到dp[i][j]的路径数，开始时dp[0][0]为1，第一行和第一列的初始化参数都为1。dp[i][j]的路径数为dp[i-1][j]的路径数和dp[i][j-1]路径数之和。

代码

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

参考文献

​​[编程题]unique-paths-ii​​