问题描述:
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.
问题解决思路:(注意题中所说的,你只能进行一次买进和卖出)
首先,对于股票来说,我们只有低价买入,高价卖出,这样才会获得最大的利润,那么反映在代码上,我们可以举一个例子
假设一个数组price={7,1,5,3,6,4} ,代表了一只股票连续六天的价格,那么,我们首先要做的就是找到最低的价格,然后我们再找到最高的价格(这里一定要注意,这里所谓的最高的价格,一定不能出现在你找到的最低的价格的前面,用这个数组来说,就是7不是最高的价格,以为时间是一去不复返的,你在第二天买进了这只股票,但你不能,再去追求昨天的价格了,得看后面的价格,这就是这个算法的核心所在,具体情况请看代码)
代码实现:
package Algorithm;
//求最大利润问题就是找到峰值和峰谷
public class Example5 {
public int Maxprofit(int[] array) {
int min = array[0];
int profit = 0;
for (int i = 1;i < array.length; i++){
if(array[i] > array[i-1]){
profit = Math.max(profit, array[i]-min);
}
else{
min = Math.min(min, array[i]);
}
}
return profit;
}
public static void main(String[] args) {
Example5 object = new Example5();
int[] array = {7,1,5,3,6,4};
System.out.print(object.Maxprofit(array));
}
}
总结:
要是新手对于这个问题不太理解的话,可以尝试画出一个折线图,然后利用小编的代码,去尝试进行每一步的操作,由慢到快,大家一起加油。另外,小编再给大家抛出一个问题,要是题中要求可以进行多次的买进和卖出,那么利润应该怎么算呢?