[leetcode] 62. Unique Paths

题目大意

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

62. Unique Paths

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).

How many possible unique paths are there?

Above is a 7 x 3 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

Example 1:

Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right

Example 2:

Input: m = 7, n = 3
Output: 28

一个机器人位于m x n隔板的左上角（在图中标记为“起点”）。

机器人在任意一点只可以向下或者向右移动一步。机器人尝试到达隔板的右下角（图中标记为“终点”）

有多少种可能的路径？

注意：m和n最多为100

解题思路

思路一：动态规划

状态转移方程：

dp[x][y] = dp[x - 1][y] + dp[x][y - 1]

初始令dp[0][0] = 1

Python代码：

class Solution(object):
    def uniquePaths(self, m, n):
        """
        :type m: int
        :type n: int
        :rtype: int
        """
        dp = [[0] * n for x in range(m)]
        dp[0][0] = 1
        for x in range(m):
            for y in range(n):
                if x + 1 < m:
                    dp[x + 1][y] += dp[x][y]
                if y + 1 < n:
                    dp[x][y + 1] += dp[x][y]
        return dp[m - 1][n - 1]

上述解法空间复杂度可以优化至O(n)：

class Solution(object):
    def uniquePaths(self, m, n):
        """
        :type m: int
        :type n: int
        :rtype: int
        """
        if m < n:
            m, n = n, m
        dp = [0] * n
        dp[0] = 1
        for x in range(m):
            for y in range(n - 1):
                dp[y + 1] += dp[y]
        return dp[n - 1]

思路二：排列组合

题目可以转化为下面的问题：

求m - 1个白球，n - 1个黑球的排列方式

公式为：

公式为：C(m + n - 2, n - 1)

Python代码：

class Solution(object):
    def uniquePaths(self, m, n):
        """
        :type m: int
        :type n: int
        :rtype: int
        """
        if m < n:
            m, n = n, m
        mul = lambda x, y: reduce(operator.mul, range(x, y), 1)
        return mul(m, m + n - 1) / mul(1, n)

参考：http://bookshadow.com/weblog/2015/10/11/leetcode-unique-paths/