其他的一些利用了分治算法的2

Question：

Find the kth largest element in an unsorted array. Note that it is the kth largest element in
the sorted order, not the kth distinct element.
INPUT: An unsorted array A and k.
OUTPUT: The kth largest element in the unsorted array A.

分析：

记该未排序的数组为A

随机生成一个数X，遍历数组A,将小于X的数放入数组A1中，大于等于X的数放入数组A2中(A1与A2仍为无序数组)，获取A1与A2的长度，记为len1与len2，若要查找的第k大的元素，其k大于len1，则要寻找的数肯定在数组A2中，则再在A2中进行如上操作，反之则在A1中进行如上操作，如果，恰巧k=len1，则随机到的这个X即为要寻找的第k大的数

伪代码

Function getKthNumber(A,len,k)
Random x
for i←0 to len do
        if A[i]<x then
            put in A1
        else put in A2
        end if
end For
len1←len(A1)
len2←len(A2)
if k=len1 then
        return x
elseif  k>len1  then
        k←k-len1
getKthNumber(A2,len2,k) 
else
getKthNumber(A1,len1,k)
end If
end Function

Question2：

Consider an n-node complete binary tree T, where n = 2d 􀀀 1 for some d. Each node v of
T is labeled with a real number xv. You may assume that the real numbers labeling the nodes
are all distinct. A node v of T is a local minimum if the label xv is less than the label xw for
all nodes w that are joined to v by an edge.
You are given such a complete binary tree T, but the labeling is only specied in the following
implicit way: for each node v, you can determine the value xv by probing the node v. Show
how to nd a local minimum of T using only O(log n) probes to the nodes of T.

分析：

从树T的根节点开始，将它的大小与自己的子节点比较，若它最小，则直接返回它，否则，看左右节点哪个小，将它当作子树的顶点，重复以上操作。

伪代码：

Function FindMin(T)
If  min(get(T.root),get(T.left),get(T.right))==get(t.root)  then
        Return  T.root
Elseif  get(T.left)<get(T.right)  then
        FindMin(T.left)
Else
        FindMin(T.right)
Endif
endFunction