Level:

​ ​ Easy

题目描述:

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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

Example 1:

Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 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
Explanation: In this case, no transaction is done, i.e. max profit = 0.

思路分析:

​​ ​ 遍历数组,保存当前遇到的最小元素minval,然后计算访问到的元素与其差值,保留遇到的最大差值,就是结果。

代码:

class Solution {
public int maxProfit(int[] prices) {
int minVal=Integer.MAX_VALUE;
int difference=0;
for(int i=0;i<prices.length;i++){
minVal=Math.min(prices[i],minVal);
difference=Math.max(difference,prices[i]-minVal);
}
return difference;
}
}

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