LeetCode——Best Time to Buy and Sell Stock II (股票买卖时机问题2)

问题： 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).

 

这道题是Best Time to Buy and Sell Stock I问题的第二个版本，题目整体的要求没有变，和I一样，只是在这里不再只限一次交易，此题中可以多次交易，只是买之前必须先卖出去，且初始手里没有股票。

 

分析： 在这道题中由于买卖的次数是不做限制的，而且是不买卖股票是不收手续费的，所以只要第二天的价格比第一天低我们就可以买，比如：

所以这道题说白了就是然你找到递增的子数列，然后所有递增子数列最大-最小的和就是最大收益。

 

代码如下（java）：

public class Solution {
public int maxProfit(int[] prices) {
if(prices == null || prices.length == 0)return 0;
int profit = 0;
for(int i=1; i<prices.length;i++){
if(prices[i]>prices[i-1]){
profit += prices[i]-prices[i-1];
}
}

return profit;
}
}