• Leetcode-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).

    Analysis:

    We calculate two things:1. from day 0 to day i, what the maximum profit we can get by performing only one transcation. 2. from any day i to the end, what the maximum profit we can get by performing one transcation. We then find out the maximum profit by adding them up for each day.

    Solution:

     1 public class Solution {
     2     public int maxProfit(int[] prices) {
     3         int len = prices.length;
     4         if (len==0 || len==1) return 0;
     5 
     6         int[] posProfit = new int[len];
     7         int[] negProfit = new int[len];
     8 
     9         int maxProfit = 0;
    10         int curProfit = 0;
    11         int minPrice = prices[0];
    12         for (int i=0;i<len;i++){
    13             if (minPrice>prices[i]) minPrice = prices[i];
    14             curProfit = prices[i]-minPrice;
    15             if (curProfit>maxProfit) maxProfit = curProfit;
    16             posProfit[i] = maxProfit;
    17         }
    18 
    19         maxProfit = 0;
    20         curProfit = 0;
    21         int maxPrice = prices[len-1];
    22         for (int i=len-1;i>=0;i--){
    23             if (maxPrice<prices[i]) maxPrice = prices[i];
    24             curProfit = maxPrice-prices[i];
    25             if (curProfit>maxProfit) maxProfit = curProfit;
    26             negProfit[i]=maxProfit;
    27         }
    28 
    29         int res = 0;
    30         for (int i=0;i<len;i++)
    31             if (res<posProfit[i]+negProfit[i]) res = posProfit[i]+negProfit[i];
    32 
    33         return res;
    34     }
    35 }
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  • 原文地址:https://www.cnblogs.com/lishiblog/p/4120725.html
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