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  • 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 };
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  • 原文地址:https://www.cnblogs.com/zhuli19901106/p/3514575.html
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