121 Best Time to Buy and Sell Stock [Leetcode]

题目内容：

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.

解题思路：
很直观，因为要使受益最大，只要找出相减最大的两个值。最大值一定要出现在最小值的后面。

解法1：
扫描一遍，记录下当前的最小值和最大值。如果最大值有更新，那么更新最大值后更新结果差。如果最小值有更新，重置最大值，继续扫描。

class Solution {public:    int maxProfit(vector<int>& prices) {        int size(prices.size()), result(0);        if(size == 0)            return result;        int min(prices[0]), max(prices[0]);        for(int i = 1; i < size; ++i) {            if(prices[i] < min) {                min = prices[i];                max = INT_MIN;            }            else if(prices[i] > max) {                max = prices[i];                int temp(max - min);                result = result > temp ? result : temp;            }        }        return result;    }};

解法2：
如果把相邻两天的股票价格相减，可以得到两天之间股票的差价。如果这个差价连续若干天相加的和达到最大，那么合适的买卖的时机。
缺点：
需要额外空间，需要扫描两遍。

class Solution {public:    int maxProfit(vector<int>& prices) {        if(prices.size() <= 1)            return 0;        vector<int> change(prices.size() - 1);        for(int i = 0; i < change.size(); ++i) {            change[i] = prices[i+1] - prices[i];        }        int result(INT_MIN), temp(0);        for(int i = 0; i < change.size(); ++i) {            temp += change[i];            if(temp < 0)                temp = 0;            else                result = result > temp ? result : temp;        }        return result > 0 ? result : 0;    }};