LeetCode -- Search in Rotated Sorted Array II

题目描述：


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.


思路：


这个题目与Search in Rotated Sorted Array基本是一样的，在旋转过的排序好的数组中进行二分查找，区别只是在于这个题目的输入可能包含重复元素。
本题解法基本与Search in Rotated Sorted Array I相同，区别只是需要跳过重复元素。
这里是Search in Rotated Sorted Array的解法：
<link>


而对于本题，需要在得到中间索引m后跳过左右重复元素，并重新计算m：
while(l < m && nums[l] == nums[m]){
    l++;
    }
    while(r > m && nums[r] == nums[m]){
    r--;
    }
m = (l + r) / 2;






实现代码：

public class Solution {
    public bool Search(int[] nums, int target) {
        if(nums.Length == 0)
    	{
    		return false;
    	}
    	
    	if(nums.Length == 1)
    	{
    		return nums[0] == target ;
    	}
    	
    	var l = 0;
    	var r = nums.Length - 1;
    	
    	while(l < r - 1){
    	var m = (l + r) / 2;
    	while(l < m && nums[l] == nums[m]){
    		l++;
    	}
    	while(r > m && nums[r] == nums[m]){
    		r--;
    	}
	    m = (l + r) / 2;
	    
    	if(nums[m] == target || nums[r] == target || nums[l] == target){
    		return true;
    	}
    	
    	if(nums[m] > nums[l] && nums[m] > nums[r]){
    		if(target < nums[r]){
    			l = m;
    		}
    		else {
    			if(target > nums[m]){
    				l = m;	
    			}
    			else{
    				r = m;
    			}
    		}
    	}
    	else if(nums[m] < nums[l] && nums[m] < nums[r]){
    		if(target > nums[l]){
    			r = m;
    		}
    		else{
    			if(target > nums[m]){
    				l = m;	
    			}
    			else{
    				r = m;
    			}
    		}
    		
    	}
    	else if(nums[m] > nums[l] && nums[m] < nums[r]){
    		if(target > nums[m]){
    			l = m;
    		}
    		else {
    			r = m;
    		}
    	}
    	}
    	
    	if(target == nums[l] || target == nums[r]){
    		return true;
    	}
    	
    	return false;
    }
}