121. Best Time to Buy and Sell Stock

Say you have an array for which the ith 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.

Example 1:
Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)
Example 2:
Input: [7, 6, 4, 3, 1]
Output: 0

In this case, no transaction is done, i.e. max profit = 0.
思路：先扫一遍，求出差值，然后就是求最大子序列和问题

class Solution {public:    int maxProfit(vector<int>& prices) {        int size = prices.size();        int tmpsum = 0, maxsum = 0;        if(size == 0) return maxsum;        int *F = new int[size - 1];        for(int i = 1; i < size; ++i)            F[i - 1] = prices[i] - prices[i - 1];        for(int i = 0; i < size - 1; ++i){            tmpsum += F[i];            if(tmpsum < 0)                tmpsum = 0;            else {                if(tmpsum > maxsum)                    maxsum = tmpsum;            }        }        return maxsum;    }};

又做了一遍，不用这么麻烦

class Solution {public:    int maxProfit(vector<int>& prices) {        int len = prices.size();        if(len == 0) return 0;        int buy = prices[0];        int profit = 0;        for(int i = 1; i < len; ++i){            if(prices[i] < buy)                buy = prices[i];            else if(prices[i] > buy){                if(prices[i] - buy > profit)                    profit = prices[i] - buy;            }        }        return profit;    }};