股票收益最大

Best Time to Buy and Sell Stock

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• 问题描述：leetcode:121
• Say you have an array for which the ith element 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.

思路：

• 这是一个动态规划问题，是一个局部与全局的问题。
• 考虑第i天出售股票的话能赚多少钱，记为local[i];其取值可以是：

    int diff = prices[i] - prices[i-1];    local[i] = max(local[i-1]+diff, diff);

• 对于全局最大为maxans：

maxans = max(maxans, local[i]);

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我自己的渣代码：

    int maxProfit(vector<int>& prices) {        int maxans = 0;        vector<int> local(prices.size(), 0);        if (prices.size()==1||prices.size()==0)            return 0;        for(int i=1; i<prices.size(); i++){            int diff = prices[i] - prices[i-1];            local[i] = max(local[i-1]+diff, diff);            maxans = max(maxans, local[i]);        }        return maxans;    }

leetcode优秀代码：

int maxProfit(vector<int> &prices) {    int maxPro = 0;    int minPrice = INT_MAX;    for(int i = 0; i < prices.size(); i++){        minPrice = min(minPrice, prices[i]);        maxPro = max(maxPro, prices[i] - minPrice);    }    return maxPro;}

minPrice is the minimum price from day 0 to day i. And maxPro is the maximum profit we can get from day 0 to day i.

How to get maxPro? Just get the larger one between current maxPro and prices[i] - minPrice.

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