leetcode练习题二：买卖股票的最佳时机

public int (int[] prices) {
    int maxProfit = 0;
    for (int i = 0; i < prices.length - 1; i++) {
        int x = prices[i];
        //拿到每一个数，获取后面的数 算出最大差
        for (int j = i + 1; j <= prices.length - 1; j++) {
            if (prices[j] - x > maxProfit) {
                maxProfit = prices[j] - x;
            }
        }
    }
    return maxProfit;
}

//最优解法：一次遍历
   public int maxProfit2(int[] prices) {
       int maxprofit = 0;
       if (prices.length == 0) {
           return maxprofit;
       }
       int minprice = prices[0];
       for (int i = 1; i < prices.length; i++) {
           //小于最小值则赋值，否则获取差
           if (prices[i] < minprice) {
               minprice = prices[i];
           } else if (prices[i] - minprice > maxprofit) {
               //判断是否是最大差值
               maxprofit = prices[i] - minprice;
           }
       }
       return maxprofit;
   }