题目描述:

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.

Example 1:

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Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

Example 2:

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Input: [7, 6, 4, 3, 1]
Output: 0

In this case, no transaction is done, i.e. max profit = 0.

这道题我一开始看错了题意. 它的意思是只允许一次买入卖出操作或者没有操作, 所以只要找到相差最大的两个价格并且低价在高价之前就可以了. 从前往后遍历一次数组, 记录到目前为止的最低价格, 然后再记录一个差额的最大值就可以了. 时间复杂度O(n)

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class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if(prices.size() < 2) return 0;
        int lowest = prices[0];
        int profit = 0;
        for(int i = 1; i < prices.size(); i++){
            if(lowest > prices[i]){
                lowest = prices[i];
            }
            else{
                profit = max(profit, prices[i] - lowest);
            }
        }
        return profit;
    }
};