package code; /* * 122. Best Time to Buy and Sell Stock II * 题意：买卖股票最大利润，可以买多次 * 难度：Easy * 分类：Array, Greedy * 思路：计算 prices[i] 与 prices[i-1] 的差值，把正数全加起来就行了 * Tips：lc121, lc309, lc188, lc123, lc714 */ public class lc122 { public int maxProfit(int[] prices) { if(prices.length<1) return 0; int res = 0; for (int i = 1; i < prices.length ; i++) { if(prices[i]-prices[i-1]>0) res += prices[i]-prices[i-1]; } return res; } }