zoukankan      html  css  js  c++  java
  • 122. Best Time to Buy and Sell Stock II (Array;Greedy)

    Say you have an array for which the ith element 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).

    法I:从左向右扫描,找到每个递增序列的第一个元素和最后一个元素

    class Solution {
    public:
        int maxProfit(vector<int> &prices) {
            if(prices.empty()) return 0;
            if(prices.size() == 1) return 0;
            int maxProfit = 0;
            int buyPrice = 0;
            int prePrice;
            bool asc = false; //if in ascending sequence
            
            vector<int>::iterator it= prices.begin(); 
            prePrice = *it;
            it++;
            for(; it < prices.end(); it++)
            {
                if ((*it)>prePrice)
                {
                    if(!asc) //find the first element in the ascending sequence
                    {
                        buyPrice = prePrice;
                        asc = true;
                    }
                }
                else  if ((*it)< prePrice)
                {
                    if(asc) //find the last element in the ascending sequence
                    {
                        maxProfit = prePrice - buyPrice + maxProfit;
                        asc = false;
                    }
                }
                prePrice = *it;
            }
            if(asc)
            {
                maxProfit = prePrice - buyPrice + maxProfit;
            }
            return maxProfit;
        }
    };

     法II:递增就就算profit

    class Solution {
    public:
        int maxProfit(vector<int>& prices) {
            int profit = 0;
            for (int i=0; i<(int)prices.size()-1; i++) {
                if (prices[i+1] > prices[i])
                    profit += prices[i+1] - prices[i];
            }
            return profit;
        }
    };
  • 相关阅读:
    2018年3月份的PTA(一)
    【性能测试】压测接口选取标准
    【jmeter】dubbo 枚举数据类型
    【python】pymysql.err.InternalError Bad handshake
    【性能调优】高并发缓存穿透
    【性能测试】系统性能容量
    【性能调优】限流策略
    快应用开发流程
    【代码运行服务】并发运行冲突
    this.$emit和bus.$emit的区别
  • 原文地址:https://www.cnblogs.com/qionglouyuyu/p/4878340.html
Copyright © 2011-2022 走看看