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).


才意识到能够在整个区间的每一点切开,然后分别计算左子区间和右子区间的最大值,然后再用O(n)时间找到整个区间的最大值。

看来以后碰到与2相关的问题,一定要想想能不能用二分法来做!
 
以下复制pickless的解说,我认为我不能比他讲的更好了
O(n^2)的算法非常easy想到:
找寻一个点j,将原来的price[0..n-1]切割为price[0..j]和price[j..n-1],分别求两段的最大profit。
进行优化:
对于点j+1,求price[0..j+1]的最大profit时,非常多工作是反复的,在求price[0..j]的最大profit中已经做过了。
类似于Best Time to Buy and Sell Stock,能够在O(1)的时间从price[0..j]推出price[0..j+1]的最大profit。
可是怎样从price[j..n-1]推出price[j+1..n-1]?反过来思考,我们能够用O(1)的时间由price[j+1..n-1]推出price[j..n-1]。
终于算法:
数组l[i]记录了price[0..i]的最大profit,
数组r[i]记录了price[i..n]的最大profit。
已知l[i],求l[i+1]是简单的,相同已知r[i],求r[i-1]也非常easy。
最后,我们再用O(n)的时间找出最大的l[i]+r[i],即为题目所求。

package Level4;  

import java.util.Arrays;  

/**
* 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).
*
*/
public class S123 { public static void main(String[] args) {
// int[] prices = {3,3,5,0,0,3,1,4};
int[] prices = {2,1,2,0,1};
System.out.println(maxProfit(prices));
} // 基本思想是分成两个时间段,然后对于某一天,计算之前的最大值和之后的最大值
public static int maxProfit(int[] prices) {
if(prices.length == 0){
return 0;
} int max = 0;
// dp数组保存左边和右边的利润最大值
int[] left = new int[prices.length]; // 计算[0,i]区间的最大值
int[] right = new int[prices.length]; // 计算[i,len-1]区间的最大值 process(prices, left, right); // O(n)找到最大值
for(int i=0; i<prices.length; i++){
max = Math.max(max, left[i]+right[i]);
} return max;
} public static void process(int[] prices, int[] left, int[] right){
left[0] = 0;
int min = prices[0]; // 左边递推公式
for(int i=1; i<left.length; i++){
left[i] = left[i - 1] > prices[i] - min ? left[i - 1] : prices[i] - min;
min = prices[i] < min ? prices[i] : min;
} right[right.length-1] = 0;
int max = prices[right.length-1];
// 右边递推公式
for(int i=right.length-2; i>=0; i--){
right[i] = right[i + 1] > max - prices[i] ? right[i + 1] : max - prices[i];
max = prices[i] > max ? prices[i] : max;
} // System.out.println(Arrays.toString(left));
// System.out.println(Arrays.toString(right));
} }

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