Leetcode 63:Unique Paths II

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.

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//这道题跟 Unique Paths 差不多，只是这道题给机器人加了障碍，不是每次都有两个选择（向右，向下）了。//因为有了这个条件，所以 Unique Paths 中最后一个直接求组合的方法就不适用了，这里最好的解法就是用动态规划了。//递推式还是跟 Unique Paths 一样，只是每次我们要判断一下是不是障碍，如果是，则dp[i][j] = 0，//否则还是dp[i][j] = dp[i - 1][j] + dp[i][j - 1]。class Solution {public:int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {int m = obstacleGrid.size(), n = obstacleGrid[0].size();if (m== 0 || n == 0)return 0;vector<int> dp(n);dp[0] = 1;for (int i = 0; i < obstacleGrid.size(); ++i){for (int j = 0; j < obstacleGrid[0].size();++j){if (obstacleGrid[i][j] == 1)dp[j] = 0;else{if (j>0)   dp[j] = dp[j] + dp[j - 1];}}}return dp[n - 1];}};

下面还有一种更容易理解的方法：

class Solution {public:int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {int i, j, m = obstacleGrid.size(), n = obstacleGrid[0].size();vector<vector<int> > dp(m, vector<int>(n, 1));if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 1) {return 0;}//初始化动态规划矩阵dpfor (i = 0; i < m; i++){if (obstacleGrid[i][0] == 0) dp[i][0] = 1;else break;}while (i < m) //某一行有障碍物，则后面所有行都为0{dp[i][0] = 0;i++;}for (j = 0; j <n; j++) {if (obstacleGrid[0][j] == 0) dp[0][j] = 1;else break;}while (j < n) //某一列有障碍物，则后面所以列都为0{dp[0][j] = 0;j++;}//动态规划核心代码for (i = 1; i < m; i++) {for (j = 1; j < n; j++) {if (obstacleGrid[i][j] == 1) dp[i][j] = 0;else dp[i][j] = dp[i][j - 1] + dp[i - 1][j];}}return dp[m - 1][n - 1];}};