63. Unique Paths II

題目描述

Discuss
Pick One

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.

題解

使用動態規劃自底向上解決，設狀態f[i][j]為(1,1)到(i,j)的路線條數，僅能向右向下走，可得狀態轉移方程：
注意：
1. 第一列如果某一行有障礙物，後面的行全為0
2. 若起點終點有障礙，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礙物，後面的行全為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];
    }
};