81. Search in Rotated Sorted Array II

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本文介绍了一种在旋转过的有序数组中查找特定元素的方法。通过去除重复元素并利用二分查找技术，有效地解决了这一问题。文章提供了一个C++实现示例，并讨论了运行时复杂度的影响因素。

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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?

先把重复的元素去除了，然后用二分查找。

class Solution {
public:    
    bool search(vector<int>& nums, int target) {
        int len = nums.size();        
        unordered_set<int> s;
        vector<int> v;
        for(int num : nums){
            if(!s.count(num)){
                s.insert(num);
                v.push_back(num);
            }             
        }            
        int l, m, r;        
        l = 0; r = v.size() - 1;
        while(l <= r){
            m = (l + r) / 2;
            if(target == v[m])
                return true;
            else if(v[m] <= v[r]){
                if(v[m] < target && target < v[r])
                    l = m + 1;
                else if(target == v[r])
                    return true;
                else 
                    r = m - 1;
            } else if(v[l] <= v[m]){
                if(v[l] < target && target < v[m])
                    r = m - 1;
                else if(target == v[l])
                    return true;
                else 
                    l = m + 1;
            }
        }
        return false;
    }
};