[leetcode] 121. Best Time to Buy and Sell Stock

Say you have an array for which the i^th 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.

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对每一个价格，若它是卖出价，则最大的差价为此价格减去其前面所有价格中的最低价。
因此用一个变量记录计算到某一个价格是的当前最低价，遍历一次所有价格即可更新到最大差价。

我的代码：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int len = prices.size();
        if (len <= 1) return 0;
        int currentMin = prices[0];
        int currentMaxDiff = 0;
        for (int i = 1; i < len; i++) {
            if (prices[i] < currentMin) currentMin = prices[i];
            if (prices[i] - currentMin > currentMaxDiff) currentMaxDiff = prices[i] - currentMin;
        }
        return currentMaxDiff;
    }
};