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leetcode - Best Time to Buy and Sell Stock II

时间:2017-08-18 10:23:48      阅读:207      评论:0      收藏:0      [点我收藏+]

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).

分析:能够多次买卖,先记录下买入价格,假设碰到比买入价格低的。更新买入价格,碰到比买入价格高的,进行交易,同一时候更新买入价格,也是线性复杂度

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int n = prices.size();
        if(n < 2)return 0;
        
        int sum = 0;
        int buy = prices[0];
        for(int i = 1; i < n; i++)
        {
            if(prices[i] < buy)buy = prices[i];
            else if(prices[i] > buy)
            {
                sum += prices[i] - buy;
                buy = prices[i];
            }
        }
        
        return sum;
    }
};


leetcode - Best Time to Buy and Sell Stock II

原文:http://www.cnblogs.com/zhchoutai/p/7387977.html

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