leetcode--Search in Rotated Sorted Array II

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.

题意：和leetcode--Search in Rotated Sorted Array一样，但是数组中可能出现重复元素

分类：数组，二分法

解法1：如果遇到无法判断在哪边的元素，说明是重复元素，这时只能遍历查找了

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1. public class Solution {  
2.     public boolean search(int[] nums, int target) {  
3.         int low = 0;  
4.         int high = nums.length-1;      
5.         while(low<=high){  
6.             int mid = (low+high)/2;           
7.             if(nums[mid]==target) return true;  
8.             if(nums[mid]>nums[low]){//如果mid属于左边  
9.                 if(nums[mid]>target && nums[low]<=target){//如果target属于左边  
10.                     high = mid-1;  
11.                 }else{  
12.                     low = mid+1;  
13.                 }  
14.             }else if(nums[mid]<nums[low]){//如果mid属于右边  
15.                 if(target>nums[mid] && target<=nums[high]){  
16.                     low = mid+1;    
17.                 }else{  
18.                     high = mid-1;  
19.                 }  
20.             }else{  
21.                 for(int i=low;i<=mid;i++){  
22.                     if(nums[i]==target) return true;  
23.                 }  
24.                 low = mid+1;  
25.             }  
26.         }  
27.         return false;  
28.     }  
29. }  

原文链接http://blog.youkuaiyun.com/crazy__chen/article/details/46427185