[LeetCode] 63. Unique Paths II_ Medium tag: Dynamic Programming

佚名 6年前 (2019-04-23) 随笔 1992人围观抢沙发百度已收录

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).

SRE实战互联网时代守护先锋，助力企业售后服务体系运筹帷幄！一键直达领取阿里云限量特价优惠。

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

这个题目思路还是跟[LeetCode] 62. Unique Paths_ Medium tag: Dynamic Programming 类似，只不过在初始化和计算时要考虑是否为obstacle即可。

Code

class Solution:
    def uniquePath(self, nums):
        if not nums or len(nums[0]) == 0:
            return 0
        lr, lc = len(nums), len(nums[0])
        mem = [[0] * lc for _ in range(lr)]
        for i in range(lr):
            if nums[i][0] != 1:
                mem[i][0] = 1
            else:
                break
        for i in range(lc):
            if nums[0][i] != 1:
                mem[0][i] = 1
            else:
                break
        for i in range(1, lr):
            for j in range(1, lc):
                mem[i][j] = 0 if nums[i][j] == 1 else mem[i - 1][j] + mem[i][j - 1]
        return mem[lr - 1][lc - 1]