Leetcode_123_Best Time to Buy and Sell Stock III

本文是在学习中的总结，欢迎转载但请注明出处：http://blog.csdn.net/pistolove/article/details/43740415

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).

思路：

（1）题意为给定一个数组，数组中第i个元素的值对应着第i天的股票，最多只能进行两次交易，每次交易只能买入一次并卖出，求能得到的最大利润。该题为Best Time to Buy and Sell Stock和Best Time to Buy and SellStockⅡ的加强版。

（2）该题同样考查的是最大差值，但是与前面类似的两题有较大的区别。由于最多只能进行两次交易，假设在数组中存在4个点(其中数组长度大于等于4)a,b,c,d四个点使得到最值，其中a<b<=c<d，f(max)=(d-c) + (b-a)，这样可以从任意位置x将数组分为两个区间，分别为0~x和x~len-1。考虑将两个区间的值保存在两个不同的数组中，通过遍历整个数组，就得到了任意点经过两次交易的分布在两个数组中的利润值，然后遍历这两个数组，同一位置上相加得到的最大值即为所得结果。详情见下方代码。

（3）希望本文对你有所帮助。

算法代码实现：

/*** @author liqq*/
public class Solution {public int maxProfit(int[] x) {if (x == null || x.length <= 1)return 0;int[] right = new int[x.length];int[] left = new int[x.length];int rmin = x[0];for (int i = 1; i < x.length; i++) {rmin = Math.min(rmin, x[i]);right[i] = Math.max(right[i - 1], x[i] - rmin);}int lmax = x[x.length - 1];left[x.length - 1] = 0;for (int i = x.length - 2; i >= 0; i--) {lmax = Math.max(lmax, x[i]);left[i] = Math.max(left[i + 1], lmax - x[i]);}int sum = 0;for (int i = 0; i < x.length; i++) {sum = Math.max(sum, right[i] + left[i]);}return sum;}
}

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