zoukankan      html  css  js  c++  java
  • LeetCode

    Best Time to Buy and Sell Stock II

    2014.1.10 22:39

    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).

    Solution:

      The best result you could get, is that you never lose money in the stock. That is to say, always buy low and sell high, which means that you buy only when price[i] < price[i + 1], buy at price[i] and sell at price[i + 1]. That guarantees you profit every time. It's greedy.

      Time complexity is O(n), space complexity is O(1).

    Accepted code:

     1 class Solution {
     2 public:
     3     int maxProfit(vector<int> &prices) {
     4         // Note: The Solution object is instantiated only once and is reused by each test case.
     5         int res;
     6         int i, len = prices.size();
     7         
     8         res = 0;
     9         for(i = 0; i < len - 1; ++i){
    10             if(prices[i] < prices[i + 1]){
    11                 res += prices[i + 1] - prices[i];
    12             }
    13         }
    14         
    15         return res;
    16     }
    17 };
  • 相关阅读:
    Python简介
    开博啦
    关于ajax访问跨域问题
    关于PHP empty()函数的错误理解
    if判断 和&&
    元素和节点
    详解js和jquery里的this关键字
    document.body的一些用法以及js中的常见问题
    浏览器报错显示
    getAttribute()获取属性
  • 原文地址:https://www.cnblogs.com/zhuli19901106/p/3514575.html
Copyright © 2011-2022 走看看