题解 | #买卖股票的最好时机(一) 与连续子数组最大值解法一致#

import java.util.*;


public class Solution {
    /**
     * 
     * @param prices int整型一维数组 
     * @return int整型
     */
    public int maxProfit (int[] prices) {
        // write code here
        if(prices.length == 0){
            return 0;
        }
        int[] profit = new int[prices.length];
        int lowerPrice = prices[0];
        for(int i = 0; i < prices.length; i++){
            if(prices[i] > lowerPrice){
                profit[i] = prices[i] - lowerPrice;
            }
            else {
                profit[i] = 0;
                lowerPrice = prices[i];
            }
        }
        Arrays.sort(profit);
        return profit[profit.length - 1];
    }
}