[leedcode 123] 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 Solution {
public int maxProfit(int[] prices) {
/*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
定义两个数组left和right，数组left[i]记录了price[0..i]的最大profit，数组right[i]记录了price[i..n]的最大profit。
最后利用O(n)的时间求出Maxprofix = max(left(0,i) + right(i+1, n))  0<=i<n*/
int len=prices.length;
if(len<=0) return 0;
int left[]=new int[len];
int right[]=new int[len];
left[0]=0;
int minleft=prices[0];
for(int i=1;i<len;i++){
left[i]=Math.max(left[i-1],prices[i]-minleft);
if(minleft>prices[i])minleft=prices[i];
}
right[len-1]=0;
int maxright=prices[len-1];
for(int j=len-2;j>=0;j--){
right[j]=Math.max(right[j+1],maxright-prices[j]);
if(maxright<prices[j])maxright=prices[j];
}
int res=0;
for(int i=0;i<len;i++){
int temp=right[i]+left[i];
if(res<temp) res=temp;
}
return res;
}
}