【LeetCode】63. Unique Paths II

问题描述

https://leetcode.com/problems/unique-paths-ii/#/description

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: mand n will be at most 100.

算法

设f(i,j)是从(0,0)到(i,j)的路径数
同Unique Paths差不多，只是多了一个障碍物的判断

1. f(0,0) = (0,0) is obstacles ?0:1
2. f(0,j) = (0,j) is obstacles ?0:f(0,j-1), 0<j<n
3. f(i,0) = (i,0) is obstacles ?0:f(i-1,0), 0<i<m
4. f(i,j) = (i,j) is obstacles ?0:f(i-1,j) + f(i, j-1), i>=1 & j>=1

代码

        public int uniquePathsWithObstacles(int[][] obstacleGrid) {
	        int m = obstacleGrid.length;
	        if(m==0) return 0;
	        int n = obstacleGrid[0].length;
	        if(n==0) return 0;
	        int[][] f = new int[m][n];
	        f[0][0] = obstacleGrid[0][0]==1?0:1;
	        // 第一行
	        for(int j=1;j<n;j++) {
	        	f[0][j] = obstacleGrid[0][j]==1?0:f[0][j-1];
	        }
	        // 第一列
	        for(int i=1;i<m;i++) {
	        	f[i][0] = obstacleGrid[i][0]==1?0:f[i-1][0];
	        }
	        for(int i=1;i<m;i++) {
	        	for(int j=1;j<n;j++) {
	        		f[i][j] = obstacleGrid[i][j]==1?0:f[i-1][j] + f[i][j-1];
	        	}
	        }
	        return f[m-1][n-1];
	    }