【原题】 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). 【解释】 此为Best Time to Buy and Sell Stock的follow up,这里可以多次买入卖出,但是一次不能持有多支股票,求最多可以赚多少钱。 【思路】 数组中后面的元素的值和前面的值的差若大于0,则加上这个差,直至循环结束。求得的值即为目标值。
可能这里有人会疑问,每次不能持有多支股票,那么遇到递增的情况是否会不合题意。举个栗子:1, 3, 5。我们可以得到的最大价值为4,按照上面的算法刚开始得到2+2,同样也可以得到正确的结果。对于递增序列的话,上面的算法就相当于用递增子序列的最大值减去最小值。
public class Solution {
public int maxProfit(int[] prices) {
int max=0;
for(int i=1;i<prices.length;i++)
if(prices[i]-prices[i-1]>0)
max+=prices[i]-prices[i-1];
//System.out.println(max);
return max;
}
}