LeetCode 122. Best Time to Buy and Sell Stock II解答

废话不多说

先来看一下题目

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).

题目大意是给你一个数组，数组的第i个元素就是第i天的股票的价格，让你设计一个算法，使得你的盈利最大。你可以买卖股票若干次，但是在买入股票之前必须先卖出。
连一个example都没有，对于我这样没有玩过股票的实在太不友好了，我花了很多时间考虑最小值会不会有负数，股票怎么样才能算盈利，能不能一次买很多股之类的问题。

解题思路

我的方法是当后一天的股值比今天的高的时候就买入，后一天的股值比今天低的时候就卖出，达到了最后一天的时候如果股值还上升就立即卖出。代码如下：

public int maxProfit(int[] prices) {        int profit=0;        int buy=-1,sell=-1;        if(prices.length==0){            return 0;        }        for(int i=0;i<prices.length-1;i++){            if((prices[i]<=prices[i+1])){                if(buy==-1)                    buy=prices[i];                    sell=-1;                if(i==prices.length-2){                    sell=prices[i+1];                    profit+=sell-buy;                }                continue;            }            if((prices[i]>prices[i+1])&&buy!=-1){                sell=prices[i];                profit+=sell-buy;                buy=-1;                continue;            }        }        return profit;    }

更好的算法

我提交了之后，看了其他人的代码才发现，卧槽，原来是可以这么解的！

class Solution {    public int maxProfit(int[] prices) {        if (prices.length <= 1) {            return 0;        }        int pp = 0, p = 0, result = 0;        for (int i = 1; i < prices.length; i++) {            if (prices[i] > prices[i-1]) {                result += prices[i] - prices[i-1];            }        }        return result;    }}

直接将当前比前一天高的差值加起来就行了！