63. Unique Paths II

Problem

Follow up for “Unique Paths”:

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.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[  [0,0,0],  [0,1,0],  [0,0,0]]

The total number of unique paths is 2.

Note: m and n will be at most 100.

Solution

二维数组

先用二维数组实现，思路清晰。

不过，实现代码量好大…

class Solution {public:    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {        if (obstacleGrid[0][0] == 1) {            return 0;        }        int m = obstacleGrid.size();        int n = obstacleGrid[0].size();        vector<vector<int>> dp(100, vector<int>(100, 0)); //关于vector<vector<int>>初始化        for (int i = 0; i != m; ++i) {            if (obstacleGrid[i][0] == 0) {                dp[i][0] = 1;            }            else { //如果初始化边时有障碍，后续则为0                dp[i][0] = 0;                break;            }        }        for (int j = 0; j != n; ++j) {            if (obstacleGrid[0][j] == 0) {                dp[0][j] = 1;            }            else {                dp[0][j] = 0;                break;            }        }        for (int i = 1; i != m; ++i) {            for (int j = 1; j != n; ++j) {                dp[i][j] = (obstacleGrid[i][j] == 1 ? 0 : dp[i-1][j] + dp[i][j-1]);            }        }        return dp[m-1][n-1];    }};

可能是因为test cases太少的缘故

这代码跑得还挺快。

一维数组

稍等。