README_EN.md

81. Search in Rotated Sorted Array II

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Description

There is an integer array nums sorted in non-decreasing order (not necessarily with distinct values).

Before being passed to your function, nums is rotated at an unknown pivot index k (0 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,4,4,5,6,6,7] might be rotated at pivot index 5 and become [4,5,6,6,7,0,1,2,4,4].

Given the array nums after the rotation and an integer target, return true if target is in nums, or false if it is not in nums.

 

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

 

Constraints:

• 1 <= nums.length <= 5000
• -104 <= nums[i] <= 104
• nums is guaranteed to be rotated at some pivot.
• -104 <= target <= 104

 

Follow up: This problem is the same as Search in Rotated Sorted Array, where nums may contain duplicates. Would this affect the runtime complexity? How and why?

Solutions

Python3

class Solution:
    def search(self, nums: List[int], target: int) -> bool:
        l, r = 0, len(nums) - 1
        while l <= r:
            mid = (l + r) >> 1
            if nums[mid] == target:
                return True
            if nums[mid] < nums[r] or nums[mid] < nums[l]:
                if target > nums[mid] and target <= nums[r]:
                    l = mid + 1
                else:
                    r = mid - 1
            elif nums[mid] > nums[l] or nums[mid] > nums[r]:
                if target < nums[mid] and target >= nums[l]:
                    r = mid - 1
                else:
                    l = mid + 1
            else:
                r -= 1
        return False

Java

class Solution {
    public boolean search(int[] nums, int target) {
        int l = 0, r = nums.length - 1;
        while (l <= r) {
            int mid = (l + r) >>> 1;
            if (nums[mid] == target) return true;
            if (nums[mid] < nums[r] || nums[mid] < nums[l]) {
                if (target > nums[mid] && target <= nums[r]) l = mid + 1;
                else r = mid - 1;
            } else if (nums[mid] > nums[l] || nums[mid] > nums[r]) {
                if (target < nums[mid] && target >= nums[l]) r = mid - 1;
                else l = mid + 1;
            } else r--;
        }
        return false;
    }
}

C++

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

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