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;
}
}

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