标签：买入卖出问题
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). Subscribe to see which companies asked this question.
题意：股票的买入卖出。你只有两次机会买入卖出。
public class Solution {
public int maxProfit(int[] prices) {
if(prices==null || prices.length<2)return 0;
//O(n^2)哎
int sum=0;
for(int i=1;i<prices.length;i++){
int temp=getMax(prices,0,i)+getMax(prices,i+1,prices.length-1);
if(temp>sum)sum=temp;
}
return sum;
}
public static int getMax(int[] prices,int left,int right){
if(left>=prices.length)return 0;
int Min=prices[left];
int sum=0;
for(int i=left+1;i<=right;i++){
Min=Math.min(Min,prices[i]);
sum=Math.max(sum,prices[i]-Min);
}
System.out.println(sum);
return sum;
// }
// //first[i]从左往右遍历，表示到第i天时卖出能赚到的最大钱
// int[] first=new int[prices.length];
// //second[i]从右往左遍历，表示从第i天到最后一天能赚到的最大钱。本质还是分两部分计算。只不过是分开两次扫描数组
// int[] second=new int[prices.length];
// int min=prices[0];
// for(int i=1;i<prices.length;i++){
// min=Math.min(min,prices[i]);
// first[i]=Math.max(first[i-1],prices[i]-min);
// }
// // for(int i=0;i<prices.length;i++){
// // System.out.println(first[i]);
// // }
// int max=prices[prices.length-1];
// for(int i=prices.length-2;i>=0;i--){
// max=Math.max(max,prices[i]);
// second[i]=Math.max(second[i+1],max-prices[i]);
// }
// int ret=0;
// for(int i=0;i<prices.length;i++){
// //System.out.println(second[i]);
// ret=Math.max(first[i]+second[i],ret);
// }
// return ret;
}
}PS:解法一是直接以第i天为界，求两边的最大收益。但是需要O(N^2)。
优化：用俩数组，遍历两次。
Leetcode 123. Best Time to Buy and Sell Stock III JAVA语言
标签：买入卖出问题
原文地址：http://fulin0532.blog.51cto.com/6233825/1910811
