Leetcode 121. Best Time to Buy and Sell Stock

Tyan

发布于 2019-05-25 22:18:25

32700

代码可运行

文章被收录于专栏：SnailTyanSnailTyan

运行总次数：0

代码可运行

文章作者：Tyan

博客：noahsnail.com | CSDN | 简书

1. Description

2. Solution

Version 1

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

Version 2

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int max = 0;
        int min = INT_MAX;
        vector<int> current_max(prices.size());
        vector<int> current_min(prices.size());
        int max_profit = 0;
        int size = prices.size();
        for(int i = 0; i < size; i++) {
            if(prices[i] < min) {
                min = prices[i];
            }
            current_min[i] = min;
        }
        for(int i = size - 1; i >= 0; i--) {
            if(prices[i] > max) {
                max = prices[i];
            }
            current_max[i] = max;
        }
        for(int i = 0; i < size; i++) {
            int profit = current_max[i] - current_min[i];
            if(profit > max_profit) {
                max_profit = profit;
            }
        }
        return max_profit;
    }
};

Version 3

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