【leetcode】Unique Paths II（动态规划）

63. Unique Paths II

leetcode题目

题目描述

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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.

题解

使用动态规划自底向上解决，设状态f[i][j]为(1,1)到(i,j)的路线条数，仅能向右向下走，可得状态转移方程：

注意：

1. 第一列如果某一行有障碍物，后面的行全为0

2. 若起点终点有障碍，直接返回0

f[i][j] = f[i-1][j]+f[i][j-1]

Solution1:

class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();

if (obstacleGrid[0][0] == 1 | obstacleGrid[m-1][n-1] == 1) {
return 0;
}
vector<int>f(n,0);
f[0] = 1;
for (int i = 0; i < m; i++) {
// 第一列如果某一行有障碍物，后面的行全为0
f[0] = f[0] == 0 ? 0 : (obstacleGrid[i][0] ? 0 : 1);
for (int j = 0; j < n; j++) {
// f[i][j]=f[i-1][j]+f[i][j-1]
f[j] = obstacleGrid[i][j] ? 0:(f[j]+f[j-1]);

}
}
return f[n-1];
}
};