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33. Search in Rotated Sorted Array

中文文档

Description

There is an integer array nums sorted in ascending order (with distinct values).

Prior to 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, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].

Given the array nums after the rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.

 

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

Example 3:

Input: nums = [1], target = 0
Output: -1

 

Constraints:

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

 

Follow up: Can you achieve this in O(log n) time complexity?

Solutions

Python3

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

Java

class Solution {
    public int 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 mid;
            if (nums[mid] > target) {
                if (nums[mid] >= nums[r] && target < nums[l]) l = mid + 1;
                else r = mid - 1;
            } else {
                if (nums[mid] <= nums[l] && target > nums[r]) r = mid - 1;
                else l = mid + 1;
            }
        }
        return -1;
    }
}

C++

class Solution {
public:
    int 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 mid;
            if (nums[mid] > target) {
                if (nums[mid] >= nums[r] && target < nums[l]) l = mid + 1;
                else r = mid - 1;
            } else {
                if (nums[mid] <= nums[l] && target > nums[r]) r = mid - 1;
                else l = mid + 1;
            }
        }
        return -1;
    }
};

JavaScript

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var search = function (nums, target) {
  let l = 0, r = nums.length - 1;
  if (l > r) return -1;
  while (l <= r) {
    let mid = l + Math.floor((r - l) / 2);
    if (nums[mid] === target) return mid;
    else if (nums[mid] <= nums[r] && target <= nums[r] && target >= nums[mid])
      l = mid + 1;
    else if (nums[mid] >= nums[l] && target <= nums[mid] && target >= nums[l])
      r = mid - 1;
    else if (nums[mid] >= nums[r])
      l = mid + 1;
    else if (nums[mid] <= nums[l])
      r = mid - 1;
    else return -1;
  }
  return -1;
};

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