121 Best Time to Buy and Sell Stock

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博客围绕股票买卖问题展开，给定数组表示每日股价，仅允许一次买卖交易，需设计算法求最大收益。分析得出找出今日前最低股价与今日股价对比，若收益超之前最大收益则更新，可转化为动态规划，维护最大收益和最低股价两个全局变量。

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1 题目

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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

Example 1:

Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 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
Explanation: In this case, no transaction is done, i.e. max profit = 0.

2 尝试解

2.1 分析

股价每天在变，允许买卖一次，问最大收益。找出今天之前的最低股价，与今日股价作对比，如果收益大于之前的最大收益，更新最大收益，最后最大收益即为结果。转化为动态规划，需要维护两个全局变量，一个是最终结果最大收益，一个是今天之前的最低股价。如果今日股价低于最低股价，更新最低股价；如果今日股价高于最低股价，计算收益，如果超过最大收益，更新最大收益。

2.2 代码

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

3 标准解

int maxProfit(vector<int> &prices) {
    int maxPro = 0;
    int minPrice = INT_MAX;
    for(int i = 0; i < prices.size(); i++){
        minPrice = min(minPrice, prices[i]);
        maxPro = max(maxPro, prices[i] - minPrice);
    }
    return maxPro;
}