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 at most two transactions.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
[balabala] 自己的想法是先找出收益最大的一笔收益,然后从剩余区间中寻找次大的收益,但是在一些case上会失败,比如 [6,1,3,2,4,7]。为什么这种贪婪做法得不到全局最优解呢?如yb君指点,因为第一次交易确定会影响第二次交易的可选范围,子问题之间相互影响,所以不可用最优子问题叠加获取最优解。
正确的思路:至多能做两笔交易,而且交易之间没有重叠,想到可以用divide and conque. 如yb君说,需要综合使用多种算法的题目就会显得比较难。这题就是分治 + 动归 综合题。
public class Solution { // 以每一天为分界点,分别求出左边和右边能获得的最大收益,然后求出最大值 // 分治 + 动归 public int maxProfit(int[] prices) { if (prices == null || prices.length < 2) return 0; int[] left = new int[prices.length]; int low = prices[0]; int maxleft = 0; for (int i = 0; i < prices.length; i++) { left[i] = Math.max(maxleft, prices[i] - low); low = Math.min(low, prices[i]); } int[] right = new int[prices.length]; int high = prices[prices.length - 1]; int maxright = 0; for (int i = prices.length - 1; i >=0 ;i--) { right[i] = Math.max(maxright, high - prices[i]); high = Math.max(high, prices[i]); } int maxprofit = 0; for (int i = 0; i < prices.length; i++) { maxprofit = Math.max(maxprofit, left[i] + right[i]); } } ////////////////////////////////下面为不正确思路的实现////////////////////////////// //fail @ [6,1,3,2,4,7] class ProfitInfo { int buyDay; int sellDay; int profit; public ProfitInfo(int buyDay, int sellDay, int profit) { this.buyDay = buyDay; this.sellDay = sellDay; this.profit = profit; } } public int maxProfit(int[] prices) { if (prices == null || prices.length < 2) return 0; ProfitInfo profit1 = maxProfit(prices, 0, prices.length - 1); ProfitInfo profit2 = maxProfit(prices, 0, profit1.buyDay - 1); ProfitInfo profit3 = maxProfit(prices, profit1.sellDay + 1, prices.length - 1); return profit1.profit + Math.max(profit2.profit, profit3.profit); } private ProfitInfo maxProfit(int[] prices, int start, int end) { if (start >= end) // length of range <= 1 return new ProfitInfo(start, end, 0); int min = start; int buy = start; int profit = 0; int sell = start; for (int i = start + 1; i <= end; i++) { if (prices[i] < prices[min]) { min = i; } else if (prices[i] > prices[min]){ int currProfit = prices[i] - prices[min]; if (currProfit >= profit) { profit = currProfit; buy = min; sell = i; } } } return new ProfitInfo(buy, sell, profit); } // fail @ [2, 4, 1] buy不能同时承担最小价格日以及当前最大profit的购买日,两者不一定是同一个值 private ProfitInfo maxProfit(int[] prices, int start, int end) { if (start >= end) // length of range <= 1 return new ProfitInfo(start, end, 0); int buy = start; int profit = 0; int sell = start; for (int i = start + 1; i <= end; i++) { if (prices[i] < prices[buy]) { buy = i; } else if (prices[i] > prices[buy]){ int currProfit = prices[i] - prices[buy]; if (currProfit >= profit) { profit = currProfit; sell = i; } } } return new ProfitInfo(buy, sell, profit); } }