81. Search in Rotated Sorted Array II

problem description

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?

algorithm thought

相比于Search in Rotated Sorted Array，这题在数组中加入了重复的数字，导致会出现一些问题。比如left和right是一样的数，甚至是，left和mid和right都是一样的数，这就不好区分两个区间。首先使用一个循环处理，如果left和right一样，将left++。这样就能去除所有边界一样的情况。除去之后，就和前面没有重复数组是一样的处理情况了

code

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

algorithm analysis

本来是二分查找的变体，时间复杂度应该是O(lgn)，但是这里有个对left和right的预处理，最坏情况下，时间复杂度会到O(n)。