【Leetcode】Search in Rotated Sorted Array II

题目：

Follow up for "Search in Rotated Sorted Array":

What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

解题思路：该题与Search in Rotated Sorted Array的主要区别在于需要考虑A[low]与A[mid]相等、A[mid]与A[high]相等以及A[low]A[mid]A[high]三者相等的情形。若A[low]与A[mid]相等而A[mid]与A[high]不等，则搜索右边序列；若A[mid]与A[high]相等而A[low]与A[mid]不等，则搜索左边序列；若A[low]A[mid]A[high]三者相等，则在最坏情况下需要对两边的序列进行搜索。

给出递归程序：

class Solution {
public:
bool search(int A[], int n, int target) {
int high,low;
low=0;
high=n-1;
return Bsearch(A,low,high,target);
}

private:
bool Bsearch(int A[], int low, int high, int target){
if((high-low)==0){
return A[low]==target;
}else if((high-low)==1){
return (A[low]==target)||(A[high]==target);
}
int mid=(low+high)/2;

if(A[mid]==target)return true;

if(A[low]<A[mid]){
if((A[low]<=target)&&(A[mid]>target)){
return Bsearch(A,low,mid,target);
}else if(A[mid]==A[high]){
return false;
}else{
return Bsearch(A,mid,high,target);
}
}else if(A[mid]<A[high]){
if((A[mid]<target)&&(A[high]>=target)){
return Bsearch(A,mid,high,target);
}else if(A[low]==A[mid]){
return false;
}else{
return Bsearch(A,low,mid,target);
}
}else if(A[low]!=A[mid]){
return Bsearch(A,low,mid,target);
}else if(A[mid]!=A[high]){
return Bsearch(A,mid,high,target);
}else{
return Bsearch(A,low,mid,target)||Bsearch(A,mid,high,target);
}
}
};

本题还有另外一种思路，可以参考leetcode详解.pdf