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Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Example 1:
Input: [7, 1, 5, 3, 6, 4] Output: 5 max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)
Example 2:
Input: [7, 6, 4, 3, 1] Output: 0 In this case, no transaction is done, i.e. max profit = 0.
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解题分析：
主要思路就是，可以买一次，再卖一次，求出最大的买卖利润，
注意的是需要先买，再卖。不能先卖，再买。这个就是example给出的原因。
# -*- coding:utf-8 -*-
__author__ = 'jiuzhang'
class Solution(object):
def maxProfit(self, prices):
if len(prices) < 2:
return 0
maxProfit = 0
curMin = prices[0]
for i in xrange(len(prices)):
curMin = min(curMin, prices[i])
maxProfit = max(maxProfit, prices[i] - curMin)
return maxProfit
'''
测试部分
p = [7,1,5,3,6,4]
a = Solution()
print a.maxProfit(p)
'''
(LeetCode)Best Time to Buy and Sell Stock -- 求股票买卖的最大利润
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原文地址：http://blog.csdn.net/u012965373/article/details/52168638
