leetcode-121. Best Time to Buy and Sell Stock-优快云博客

本文链接：https://blog.youkuaiyun.com/myan991/article/details/62887295

本文介绍了一种算法，用于寻找给定股票价格序列中最佳的买入和卖出时机以获得最大利润。该算法通过一次遍历即可确定最大收益，并提供了解决方案的C++实现。

leetcode-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.

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

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

下面这种就是错误的，错误的原因是要用当前的值去减最小值，而不是要用最大值去减最小值

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if(prices.empty()) return 0;
        int tmin = prices[0];
        int tmax = prices[0];
        int res = 0;
        for(int i =1; i < prices.size(); ++i){
            if(prices[i] < tmin)
                tmin = prices[i];
            if(prices[i] > tmax)
                tmax = prices[i];
            res = max(res, tmax - tmin);
        }
        return res;
    }
};