zoukankan      html  css  js  c++  java
  • 面试题38:股票最大收益问题

    leetcode中股票问题指的是股票的价格处于波动当中,分别在最多购买一次、买卖任意多次、买卖k次的最大收益

    Say you have an array for which the ithelement 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.

     1 class Solution {
     2 public:
     3     int maxProfit(vector<int> &prices) {
     4         int res = 0;
     5         int buy = INT_MAX;
     6         for(int price : prices){
     7             buy = min(buy,price);
     8             res = max(res,price-buy);
     9         }
    10         return res;
    11     }
    12 };

    Say you have an array for which the ithelement is the price of a given stock on day i.

    Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

     1 class Solution {
     2 public:
     3     int maxProfit(vector<int> &prices) {
     4         int sum = 0;
     5         for(size_t i =1;i<prices.size();i++){
     6             if(prices[i] > prices[i-1]){
     7                 sum += prices[i] - prices[i-1];
     8             }
     9         }
    10         return sum;
    11     }
    12 };

    Say you have an array for which the ithelement is the price of a given stock on day i.

    Design an algorithm to find the maximum profit. You may complete at most twotransactions.

    Note:
    You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

    class Solution {
    public:
        int maxProfit(vector<int> &prices) {
            int n = prices.size();
            if(n < 2) return 0;
            vector<vector<int>> g(n,vector<int>(3,0));
            vector<vector<int>> l(n,vector<int>(3,0));
            for(int i=1;i<n;i++){
                int diff = prices[i] - prices[i-1];
                for(int j=1;j<=2;j++){
                    l[i][j] = max(g[i-1][j-1] +max(diff,0),l[i-1][j] + diff);
                    g[i][j] = max(g[i-1][j],l[i][j]);
                }
            }
            return g[n-1][2];
        }
    };
  • 相关阅读:
    Ionic2开发环境搭建
    关于在浏览器中测试cordova plugin的注意事项。
    像azure一样桌面显示Windows系统信息
    吐槽下微软
    Hyper-v UBUNTU 12.04 模板设置
    MongoDB整库备份与还原以及单个collection备份、恢复方法
    nginx 配置
    vsftpd配置文件详解
    分享几个免费IP地址查询接口(API)
    黑客帝国效果
  • 原文地址:https://www.cnblogs.com/wxquare/p/6944069.html
Copyright © 2011-2022 走看看