LeetCode 81. Search in Rotated Sorted Array II

Binary Search

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?

Solution

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中文版解答Bilibili Link

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