LeetCode 0081 - Search in Rotated Sorted Array II

Reck Zhang

发布于 2021-08-11 10:50:41

2060

发布于 2021-08-11 10:50:41

文章被收录于专栏：Reck Zhang

Search in Rotated Sorted Array II

Desicription

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.

Solution

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        int left = 0;
        int right = nums.size() - 1;
        while(left <= right){
            int mid = (left + right) >> 1;
            if(nums[mid] == target)
                return 1;
            while(nums[left] == nums[mid] && nums[right] == nums[mid])
                left++, right--;
            if(nums[mid] >= nums[left]){
                if(nums[left] <= target && nums[mid] > target)
                    right = mid - 1;
                else
                    left = mid + 1;
            }
            else{
                if(nums[right] >= target && nums[mid] < target)
                    left = mid + 1;
                else
                    right = mid - 1;
            }
        }
        return 0;
    }
};

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原始发表：2017-11-20，如有侵权请联系 cloudcommunity@tencent.com 删除

duplicates