Best Time to Buy and Sell Stock

题目

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.

解析

只需要遍历一次数组，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最为利润，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：

思路

1. using pointer named price to sweep the whole array, tracking the min value initially from prices[0]
2. calculating the difference between current price and min value as the profit
3. so maximum profit is either current profit or previous profit, whichever is larger

图解

代码

class Solution {
    public int maxProfit(int[] prices) {
        if(prices.length<2) return 0;
        int profit = 0;
        int min = prices[0];
        for(int price : prices){
            min = Math.min(min, price);
            profit = Math.max(profit, price - min);
        }
        return profit;  
    }
}