(DP 线性DP 递推) leetcode 63. Unique Paths II

佚名 6年前 (2019-05-10) 随笔 1781人围观抢沙发百度已收录

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).

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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

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这个是leetcode 62. Unique Paths的拓展题。其中不同点就是这个题有障碍。所以，面对这个障碍是要避开。

C++代码：

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        if(obstacleGrid[0][0] == 1 || obstacleGrid.size() == 0 || obstacleGrid[0].size() == 0) return 0;   //如果出发的贷方有障碍，当然不能出发了。
        int m = obstacleGrid.size(),n = obstacleGrid[0].size();
        vector<vector<long> > dp(m+1,vector<long>(n+1,0));
        for(int i = 0; i < m; i++){
            if(obstacleGrid[i][0] == 1)
                break;
            dp[i][0] = 1;
        }
        for(int j = 0; j < n; j++){
            if(obstacleGrid[0][j] == 1)
                break;
            dp[0][j] = 1;
        }
        for(int i = 1; i < m; i++){
            for(int j = 1; j < n; j++){
                if(obstacleGrid[i][j] == 1)
                    continue;
                dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }
        return dp[m-1][n-1];
    }
};