Leetcode - Best Time to Buy and Sell Stock III

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]。 失败的原因是获取最大收益的区间如果能拆分成两次交易则收益可能更大，如例子所示，但并不是特别理解，还请路过的大神指点一二~  
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);
    }
}