【Leetcode】Search in Rotated Sorted Array II

从一个旋转的排序数组中寻找一个数字，数组中可能有重复数字，要求时间复杂度O(LogN)。

 

( 1 1 1 2 4  might become 1 1 2 4 1).

You are given a target value to search. If found in the array return true, otherwise return false.

思路：二分查找，需要注意的是，若中间数字跟最左端数字相等时，去掉最左端的数字，然后递归查找。代码如下：

 

class Solution
{
	bool search(int A[], int n, int target) 
	{
		return search(A,0,n-1,target);
	}

	bool search(int *ary, int start, int end, int target)
	{
		if (start > end) return false;

		int mid = start + ((end-start)>>1);
		int number0 = ary[start];
		int number = ary[mid];
		int number1 = ary[end];

		if (number == target) 
			return true;

		if (number > number0)
		{
			if(target < number0 || target > number)
				return search(ary,mid+1,end,target);
			else
				return search(ary,start,mid-1,target);
		}
		else if(number < number0)
		{
			if (target > number1 || target < number)
				return search(ary,start,mid-1,target);
			else
				return search(ary,mid+1,end,target);
		}
		// number == number0
		return search(ary,start+1,end,target);
	}
};