8. 33. Search in Rotated Sorted Array

33. Search in Rotated Sorted Array

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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 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 an ascending array that is possibly rotated.

• 104 <= target <= 104

Search in Rotated Sorted Array - LeetCodeLeetCode

time: O(n)

space: O(1)

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var search = function(nums, target) {
    
    let s = 0;
    let e = nums.length - 1;
    while (s < e) {
        const m = s + Math.floor((e - s) / 2);
        if (nums[m] <= nums[e]) {
            e = m;
        } else {
            s = m + 1;
        }
    }

    const pivot = e;
    if (nums[pivot] <= target && target <= nums[nums.length - 1]) {
        s = pivot - 1;
        e = nums.length;
    } else {
        s = -1;
        e = pivot;
    }
    // 0,1,2,2,2,3,4,5,6
    while (s + 1 < e) {
        const m = s + Math.floor((e - s) / 2);
        if (target <= nums[m]) {
            e = m;
        } else {
            s = m;
        }
    }

    return nums[e] === target ? e : -1;
};