Unique Paths

题目描述：

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?

3 * 7的example

一开始无脑了一下，决定用递归：

a[m,n]=a[m-1,n] + a[m,n-1]

class Solution(object):
	def uniquePaths(self, m, n):
		if m == 1 or n == 1:
			return 1
		return self.uniquePaths(m,n-1) + self.uniquePaths(m-1,n)

自然是超时

后决定迭代计算：有两种顺序，先从左到右遍历一行，再从上到下遍历每一行，最左边初始化为1，其余的元素为其左和其上元素之和。这样细看空间需要m*n，想到了简化空间，我们容易知道输入m,n的次序不影响结果，我们可以只记录h = min(m,n)个值，我们假定高度值为h，也就是上图中对应的3，假设中间值存在数组tmp中，那么tmp[0]是1并保持不变，以后更新的时候将tmp[i] = tmp[i](旧值，相当于左侧元素）+ tmp[i-1](新值，相当于上侧元素）

class Solution(object):
	def uniquePaths(self, m, n):
		h = min(m,n)
		w = max(m,n)
		tmp = [1] * h
		for i in range(1,w):
			for j in range(1,h):
				tmp[j] = tmp[j]+tmp[j-1]

		return tmp[h-1]