Leetcode 33 - Search in rotated sorted array

Note:

  • Update
    • Don’t just compare nums[middle] with target because target can be on either side.
    • Compare nums[middle] with nums[left] first to check which situation we’re talk about.
  • Because of the time complexity requirement, we can know we must use binary search
  • The direction of moving left, right pointers depends on where the pivot is, there are two situations:
    • img
    • We can check the pos of pivot by comparing left with middle.
  • Only by checking nums[middle] > target is not enough.
    • For example: img
    • Both of these two pos of target meet nums[m] > target.
    • We need to compare target with left/right as well.

Question:

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

Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= 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 possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.

You must write an algorithm with O(log n) runtime complexity.

Example:

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

Code:

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/**
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
var search = function(nums, target) {
let left = 0;
let right = nums.length - 1;
while (left <= right) {
if (nums[right] === target) return right;
if (nums[left] === target) return left;
const middle = (left + right) >> 1;
if (target === nums[middle]) {
return middle;
}
if (nums[left] >= nums[middle]) {
if (nums[middle] < target && target < nums[right]) {
left = middle + 1;
} else {
right = middle - 1;
}
} else {
if (target < nums[middle] && target > nums[left]) {
right = middle - 1;
} else {
left = middle + 1;
}
}
}
return -1;
};