LeetCode 81. Search in Rotated Sorted Array II-优快云博客

本文链接：https://blog.youkuaiyun.com/EasonDongH/article/details/98582841

本文探讨了在一个未知旋转点的升序数组中搜索目标值的问题，提供了详细的算法实现，并讨论了包含重复元素时的时间复杂度影响。

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81. Search in Rotated Sorted Array II
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).

You are given a target value to search. If found in the array return true, otherwise return false.

Example 1:

Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true
Example 2:

Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false
Follow up:

This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
Would this affect the run-time complexity? How and why?

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        int L = 0, R = nums.size()-1, mid = 0;
        while(L <= R) {
            mid = L + (R-L)/2;
            if(nums[mid] == target)
                return true;
            if(nums[L] < nums[mid]) {
                if(nums[L]<=target && target<nums[mid])
                    R = mid - 1;
                else
                    L = mid + 1;
            }
            else if(nums[L] == nums[mid])
                L += 1;
            else {
                if(nums[mid]<target && target<=nums[R])
                    L = mid + 1;
                else
                    R = mid - 1;
            }
        }
        return false;
    }
};

这使得时间复杂度最坏情况为O(n)，因为当nums[L]==nums[R]==nums[mid]时，没法二分数组，只能线性地缩小数组范围，这在最坏情况下将是O(n)的。例如，序列[1,1,1,1,1,1]中找2，需要判断5次。