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题目
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.
方法
和同一时候找一个数组中的最大值和最小值类似。
加了限制条件:最小值仅仅能在最大值的前面。
1. 先初始化最小值和最大值为数组的第一个元素,maxProfit = 0;
2.遍历一遍数组:假设元素大于最大值,则更新最大值和maxProfit
假设元素小于最小值,同一时候更新最大值和最小值。
public int maxProfit(int[] prices) { if (prices == null) { return 0; } int len = prices.length; if (len == 0 || len == 1) { return 0; } int maxPro = 0; int min = prices[0]; int max = prices[0]; for (int i = 1; i < len; i++) { if (prices[i] > max) { max = prices[i]; int tempMaxPro = max - min; if (tempMaxPro > maxPro) { maxPro = tempMaxPro; } } else if (prices[i] < min) { min = prices[i]; max = prices[i]; } } return maxPro; }
Best Time to Buy and Sell Stock
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原文地址:http://www.cnblogs.com/bhlsheji/p/4390883.html