【Leetcode】81. Search in Rotated Sorted Array II

Question：

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

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

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

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

The array may contain duplicates.

Tips：

给定一个数组，该数组是由一个有序的数组经过旋转（即将前面一段数字接到整个数组之后）得到的。判断target是否存在于该数组之中。

本题为33题升级版本，数组中的数字可以出现重复，如果target存在，返回他的true不存在则返回false。

思路：

本题与33提相似，但是由于数组中存在重复数字，可能会出现low high mid三个数字均相等的情况，这时为了跳出相等数字，需要low++或者high--;

代码：

 public boolean search(int[] nums, int target) {
        if (nums == null)
            return false;
        int low = 0;
        int len = nums.length;
        int high = len - 1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (nums[mid] == target)
                return true;
            if (nums[low] < nums[mid] || nums[mid]>nums[high]) {
                if (target < nums[mid] && target >= nums[low]) {
                    high = mid - 1;
                } else
                    low = mid + 1;
            }else if(nums[mid]<nums[high] || nums[low]>nums[mid]){
                if(target<=nums[high] && target>nums[mid]){
                    low=mid+1;
                }else{
                    high=mid-1;
                }
            }
            else{
                low++;
            }
        }
        return false;
    }