Leetcode NO.63 Unique Paths II

题目要求如下：

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.

这题我开始没有做出来，因为原来我虽然学过数据结构，但是那门课覆盖的知识并不全，没有介绍dynamic programming算法。。而之前dp的题过于简单，即便不用dp的思想，我也能做出来，但是这题就不行了。。

简要介绍下本题的主干思路：

上图矩阵为matrix[3][3]，再设定一个dp[3][3]矩阵

dp[i][j]就相当于matrix[0][0]到matrix[i][j]的路径数目。

而所谓dynamic programming，就是d[i][j] = d[i-1][j] + d[i][j-1]。。有点类似于斐波那契数列。

则最后返回的值就是dj[2][2]。

我看讨论区有一种更加简便的方法，但是我认为这么做更有利于我理解dp的思想，所以就没用那种更好的算法。。

详细代码如下所示：

class Solution {
public:int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {int height = obstacleGrid.size();int width = obstacleGrid[0].size();if (height == 0 or width == 0)return 0;if (obstacleGrid[0][0] == 1 or obstacleGrid[height-1][width-1] == 1)return 0;vector<vector<int>> dp(height, vector<int>(width, 0));dp[0][0] = 1;/* set the first row */for (int i = 1; i < width; ++i) {if (obstacleGrid[0][i] == 0)dp[0][i] = dp[0][i-1];}/* set the first column */for (int j = 1; j < height; ++j) {if (obstacleGrid[j][0] == 0)dp[j][0] = dp[j-1][0];}/* other part */for (int i = 1; i < width; ++i) {for (int j = 1; j < height; ++j) {if (obstacleGrid[j][i] == 0)dp[j][i] = dp[j-1][i] + dp[j][i-1];}}return dp[height-1][width-1];}
};

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