Leetcode 81 - Search in rotated sorted array II

Note:

  • Find the middle = left + right / 2.
  • But because it has duplicates, when nums[left] == nums[right] == nums[middle],
    • Such as [1,1,1,2,1,1,1,1,1,1,1] or [1,1,1,1,1,1,1,1,2,1,1] with target 2.
    • There is no way we can tell 2 is on the left of middle or right.
  • What should we do?
    • Move both left and right by 1.
  • Use nums[midlle] <= nums[right] to check which part the middle is in.
  • We should use <= as the condition of while so that we don’t need an extra check on nums[left] === target.

Question:

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.

You must decrease the overall operation steps as much as possible.

Example:

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

Code:

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