【LeetCode】63. Unique Paths II

内容简介：【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: m and 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];
        }

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：

http://www.zgljl2012.com/leetcode-63-unique-paths-ii/

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