best-time-to-buy-and-sell-stock-i
Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.
public class maxProfit { public int maxProfit(int[] prices) { if (prices == null || prices.length == 0 ) return 0; int min = prices[0]; int maxprofit = 0; for (int i=0;i<prices.length;i++) { if (prices[i] < min) { min = prices[i]; } if (prices[i] - min >maxprofit) { maxprofit = prices[i] - min; } } return maxprofit; } }
best-time-to-buy-and-sell-stock-ii
Say you have an array for which the i th 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).
给定一个数组,它的第 i 个元素是一支给定股票第 i 天的价格。
设计一个算法来计算你所能获取的最大利润。你可以尽可能地完成更多的交易(多次买卖一支股票)。
注意:你不能同时参与多笔交易(你必须在再次购买前出售掉之前的股票)。
示例 1:
输入: [7,1,5,3,6,4] 输出: 7 解释: 在第 2 天(股票价格 = 1)的时候买入,在第 3 天(股票价格 = 5)的时候卖出, 这笔交易所能获得利润 = 5-1 = 4 。 随后,在第 4 天(股票价格 = 3)的时候买入,在第 5 天(股票价格 = 6)的时候卖出, 这笔交易所能获得利润 = 6-3 = 3 。
判断相邻是否递增,因为连续递增可以合起来看为一次买入卖出操作,所以统计所有递增量即可
public class bestTimeToBuyAndSellStockii { public int maxProfit(int[] prices) { if (prices == null || prices.length == 0 ) return 0; int profit = 0; for (int i=0;i<prices.length-1;i++) { if (prices[i] < prices[i+1]) { profit += prices[i+1] - prices[i]; } } return profit; } }