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?

 

AC code:

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

Runtime: 4 ms, faster than 99.15% of C++ online submissions for Search in Rotated Sorted Array II.

 

题解：找出排列有序的一端进行二分查找。