149. Best Time to Buy and Sell Stock【medium】

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

Given array [3,2,3,1,2], return 1.

 

题意

假设有一个数组，它的第i个元素是一支给定的股票在第i天的价格。若是你最多只容许完成一次交易(例如,一次买卖股票),设计一个算法来找出最大利润。

 

解法一：

 1 class Solution {
 2 public:
 3     /**
 4      * @param prices: Given an integer array
 5      * @return: Maximum profit
 6      */
 7     int maxProfit(vector<int> &prices) {
 8         if (prices.empty()) {
 9             return 0;
10         }
11         int i = prices.size() - 1;
12         int ans = 0;
13         int maxp = prices[i];
14         for (--i; i >= 0; --i){
15             ans = max(ans, maxp - prices[i]);
16             maxp = max(maxp, prices[i]);
17         }
18         return ans;
19     }
20 };

参考@NineChapter 的代码

 

解法二：

 1 public class Solution {
 2     public int maxProfit(int[] prices) {
 3         if (prices == null || prices.length == 0) {
 4             return 0;
 5         }
 6 
 7         int min = Integer.MAX_VALUE;  //just remember the smallest price
 8         int profit = 0;
 9         for (int i : prices) {
10             min = i < min ? i : min;
11             profit = (i - min) > profit ? i - min : profit;
12         }
13 
14         return profit;
15     }
16 }

参考@NineChapter 的代码