LeetCode 121. Best Time to Buy and Sell Stock

MaSay 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.

本题目其实就是寻找两个数的最大差值，最朴素的思想就是两个指针遍历，时间复杂度O(N2) 。

 public class Solution {    public int maxProfit(int[] prices) {        int maxProfit = 0;        for(int i= 0 ; i<prices.length; i++){            for(int j =i+1; j<prices.length;j++){                int profit = prices[j]-prices[i];                profit = profit >0?profit:0;                maxProfit = maxProfit >profit? maxProfit: profit;            }        }        return maxProfit;    }}

很不幸，和我心中想的一样，超时了。
参考别人的思路，发现其实可以做到O(N)的时间复杂度。

public class Solution {    public int maxProfit(int[] prices) {        if(prices.length ==0){            return 0;        }        int maxProfit = 0;        int curMin = prices[0];        for(int i= 1 ; i<prices.length; i++){            curMin = curMin > prices[i]? prices[i]:curMin;            maxProfit = prices[i]-curMin> maxProfit? prices[i]-curMin: maxProfit;        }        return maxProfit;    }}