对于给出的一组数据，要找到它们的最大或者最小值，运用递归和分治的思想方法来解决问题。n=2时，一次比较就可以找出两个数据元素的最大元和最小元。 当n>2时，可以把n个数据元素分为大致相等的两半， 一半有?n/2?个数据元素，而另一半有?n/2?个数据元素。 先分别找出各自组中的最大元和最小元，然后 将两个最大元进行比较，就可得n个元素的最大元； 将两个最小元进行比较，就可得n个元素的最小元。-For a given set of data to find their maximum or minimum, the use of recursion and divided way of thinking to solve the problem. n = 2, the first comparison of two data elements can be found on the maximum and minimum yuan yuan. When n>　2 when n data elements can be divided into roughly equal halves, half of the? N/2? Data elements, while the other half there? N/2? Data elements. First, to identify their group were the largest and smallest yuan yuan, and then compare the two greatest element, you can have n elements of the greatest element　to compare the two smallest element, you can get the smallest element of n-yuan .