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.

/** 旋转数组中寻找target，主要考虑左、右边界与中间元素相等
3 1 2 3 3 3 3
3 3 3 3 1 2 3    这种情况无法排除掉保留那一半数据，只能去掉首尾元素缩进，退化到O(n)复杂度
**/
 
class Solution {
public:
    bool search(int A[], int n, int target)
    {
     int low = 0,high = n-1;
    while(low<=high)
    {
       int middle = (low+high)>>1;
       if(A[middle] == target)
          return true;
       if(A[middle]==A[high] && A[middle]==A[low])    /// 比unique查找多了两句
       {
           low++;
           high--;
       }
       else if(A[low] <= A[middle]) /// left side is sorted
       {
            if(A[low]<=target && target<A[middle])
                high = middle-1;
            else
                low = middle+1;
       }
       else                       /// right side id sorted
       {
            if(A[middle]<target && target<=A[high])
                low = middle+1;
            else
                high = middle-1;
       }
    }
    return false;
    }
};