第五周：[Leetcode]121. Best Time to Buy and Sell Stock

原创于 2017-03-26 00:18:22 发布 · 318 阅读

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本文介绍了一种算法，用于计算给定股票价格序列中能够获得的最大利润。该算法通过一次遍历找到买入和卖出的最佳时机，确保交易合法性。示例展示了如何应用此算法解决实际问题。

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:
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:
Input: [7, 6, 4, 3, 1]
Output: 0
In this case, no transaction is done, i.e. max profit = 0.

寻找数组中最大值与最小值之差，最小值的下标应该在最大值之前。使用一个变量维护当前数组的最小值，向后遍历，寻找最大的差值。

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