121. Best Time to Buy and Sell Stock

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.
含义：假设有一个数组，它的第i个元素是一个给定的股票在第i天的价格。设计一个算法来找到最大的利润。你最多可以完成一次交易。
思路：遍历一次数组，用一个变量记录遍历过数中的最小值，然后每次计算当前值和这个最小值之间的差值做为利润，然后每次选较大的利润来更新。当遍历完成后当前利润即为所求，代码如下：

public class Solution {
    public int maxProfit(int[] prices) {
        int res = 0, buy = Integer.MAX_VALUE;
        for (int price : prices) {
            buy = Math.min(buy, price);
            res = Math.max(res, price - buy);
        }
        return res;
    }
}

 

 类似于53. Maximum Subarray