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63. Unique Paths II

时间:2018-10-21 14:17:22      阅读:175      评论:0      收藏:0      [点我收藏+]

标签:分享   not   cti   problems   c++   bottom   tom   with   ems   

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?

技术分享图片

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

AC code:

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

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Unique Paths II.

 

63. Unique Paths II

标签:分享   not   cti   problems   c++   bottom   tom   with   ems   

原文地址:https://www.cnblogs.com/ruruozhenhao/p/9824784.html

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