[leetcode] 63. Unique Paths II 解题报告

题目链接：https://leetcode.com/problems/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.

思路: 和上一个一样都是动态规划的题目，只是稍微有点不同，就是多了不可以走的格子，所以不可以走的地方就直接可以设为到这里的路径为0。另外需要注意的是初始化状态的时候第一行和第一列，只要出现一个1，那么接下来的都将为0，因为上一个是不可到达的，而到下一个又必须从他经过。时间复杂度为O(M*N)，空间复杂度为O(M*N)。具体代码如下：

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