LeetCode(63)-Unique Paths II

63. Unique Paths II

A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).
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.
Note: m and n will be at most 100.
Example 1:
Input:
[
[0,0,0],
[0,1,0],
[0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:

  1. Right -> Right -> Down -> Down
  2. Down -> Down -> Right -> Right

嗯?这个题还不会做,这是一个大神的代码,暂时还看不懂。明天再看

public int uniquePathsWithObstacles(int[][] obstacleGrid) {
    int width = obstacleGrid[0].length;
    int[] dp = new int[width];
    dp[0] = 1;
    for (int[] row : obstacleGrid) {
        for (int j = 0; j < width; j++) {
            if (row[j] == 1)
                dp[j] = 0;
            else if (j > 0)
                dp[j] += dp[j - 1];
        }
    }
    return dp[width - 1];
}
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值