Best Time to Buy and Sell Stock III -- Leetcode

12.25 2014

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).

class Solution {public:    int maxProfit(vector<int> &prices) {        int day=prices.size();        if(day<2) return 0;        int Max=0;        int Max1=0;        int Max2=0;        int lowest=prices[0];        for(int i=0;i<day;i++){            Max1=helper(prices,0,i);            Max2=helper(prices,i+1,day-1);            Max=max(Max1+Max2,Max);        }        return Max;    }    int helper(vector<int> &prices, int start, int end){        int profit=0;        int lowest=prices[start];        for(int i=start;i<=end;i++){            profit=max(prices[i]-lowest,profit);            lowest=min(lowest,prices[i]);        }        return profit;    }};

做法一：拆成两段，对两段分别求最大和，加起来的和取最大值即为答案。但是会出现超时。

class Solution {public:    int maxProfit(vector<int> &prices) {        int day=prices.size();        if(day<2) return 0;        vector<int> v(day,0);        vector<int> p(day,0);        for(int i=1, valley=prices[0];i<day;i++){            valley=min(valley,prices[i]);            v[i]=max(v[i-1],prices[i]-valley);        }        for(int i=day-2, peak=prices[day-1];i>=0;i--){            peak=max(peak,prices[i]);            p[i]=max(p[i],peak-prices[i]);        }        int profit=0;        for(int i=0;i<day;i++)            profit=max(profit,v[i]+p[i]);        return profit;    }};

做法二：

稍微有些不同，主要是在代码上的区别，空间和时间上的tradeoff，把各种可能先算一遍存在vector内然后再进行求最大值。