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[LeetCode]Best Time to Buy and Sell Stock II

时间:2017-08-12 15:11:29      阅读:221      评论:0      收藏:0      [点我收藏+]

题目描写叙述


Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

给你一个数组。array[i]表示第i天股票的买入卖出价格,你能够无限次买入卖出。但不能同一时候进行。求最大收益。

解题思路


实际上是折线图中全部上升段的长度和。

即求每一个子递增序列的最大最小差值的和。


代码


public static int maxProfit(int[] prices) {
		int max = 0;
		int i = 0;

		if (prices.length < 2) {
			return 0;
		}

		while (i < prices.length - 1) {
			int base = prices[i];
			while (i < prices.length - 1 && prices[i + 1] >= prices[i]) {
				i++;
			}
			max += prices[i] - base;
			i++;
		}

		return max;
	}


[LeetCode]Best Time to Buy and Sell Stock II

原文:http://www.cnblogs.com/liguangsunls/p/7350329.html

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