买卖股票的最佳时机

public class Solution {

    /**

     * @param prices: Given an integer array

     * @return: Maximum profit

     */

    public int maxProfit(int[] prices) {

        // write your code here

        int max_pro = 0;

      int temp;

     for (int m = 0; m < prices.length-1; m++) {

         for (int n = m + 1; n <prices.length; n++) {

             if (prices[m] > prices[n])

                 continue;

             else

                temp = prices[n] - prices[m];

             if(max_pro<temp)

                 max_pro = temp;

        }

    }

 

     return max_pro;

     }

 }

posted on 2017-03-14 18:06  TYYD  阅读(93)  评论(0编辑  收藏  举报

导航