[leetcode] Best Time to Buy and Sell Stock

Problem:

Say you have an array for which the i^th element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Solution1:

class Solution {public:    int maxProfit(vector<int> &prices) {        int max = 0;        for(int i = 0; i<prices.size(); i++){            for(int j = i ; j<prices.size(); j++){                int curProfit = prices[j] - prices[i];                if(max < curProfit){                    max = curProfit;                }            }        }        return max;    }};

最简单的思路，遍历两遍，复杂度O(n²)，of course超时了。

Solution2:

可以采用动态规划的方法。但是用动态规划的算法，都会产生二位数组来保存中间状态，时间复杂度不会降低。

突然想到可以转换为求连续子数组的最大和的问题，复杂度就是O(n)了。

即计算前一个数和后一个数的差值，求差值数组中最大连续子数组的和。轻轻松松就Accepted了。

 int maxProfit(vector<int> &prices) {      int max = 0;    int curProfit = 0;    if(prices.size()==0){        return max;    }    vector<int> profits;    for(int i = 0; i < prices.size()-1; i++){        int profit = prices[i+1] - prices[i];        profits.push_back(profit);    }    for(int i = 0; i<profits.size(); i++){        if(curProfit + profits[i] > 0){            curProfit = curProfit + profits[i];            if(curProfit > max){                max = curProfit;            }        }else{            curProfit = 0;                                                                                                                           }    }    return max;    }