LeetCode——Best Time to Buy and Sell Stock III

Description：

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).

给出股价表，最多两次交易，求出最大收益。

动态规划，分解成两个子问题，在第i天之前（包括第i天）的最大收益，在第i天之后（包括第i天）的最大收益，这样就变成了求一次交易的最大收益了，同系列问题I，

最后遍历一次求子问题最大收益之和的最大值，就是最后的解。时间复杂度O(n)，空间复杂度O(n)

public class Solution {
    public int maxProfit(int[] prices) {
        int len = prices.length;

        if(len == 0) return 0;

        int[] leftProfit = new int[len];
        int[] rightProfit = new int[len];

        Arrays.fill(leftProfit, 0);
        Arrays.fill(rightProfit, 0);

        //计算左边的最大收益，从前向后找
        int lowestPrice = prices[0];
        for(int i=1; i<len; i++) {
            lowestPrice = min(lowestPrice, prices[i]);
            leftProfit[i] = max(leftProfit[i-1], prices[i] - lowestPrice);
        }

        //计算右边的最大收益，从后往前找
        int highestPrice = prices[len-1];
        for(int i=len-2; i>=0; i--) {
            highestPrice = max(highestPrice, prices[i]);
            rightProfit[i] = max(rightProfit[i+1], highestPrice-prices[i]);
        }

        int maxProfit = leftProfit[0] + rightProfit[0];
        //遍历找出经过最多两次交易后的最大了收益
        for(int i=1; i<len; i++) {
            maxProfit = max(maxProfit, leftProfit[i] + rightProfit[i]);
        }

        return maxProfit;
    }

    public int min(int a, int b) {
        return a > b ? b : a;
    }

    public int max(int a, int b) {
        return a > b ? a : b;
    }
}