LeetCode 33. Search in Rotated Sorted Array

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 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

Source: https://leetcode.com/problems/search-in-rotated-sorted-array/

Solution

The main idea is first using binary search to find the pivot and then performing another binary search on a part of the array.

For the binary search in findPivot, left must be adjacent to right (not overlapped). Because we use a conservative strategy to move pointers (left = mid, right = mid).

// find the pivot of rotated sorted array
// that divide nums into two interval: [:pivot), [pivot:]
// @pre: nums only contains distinct numbers
private int findPivot(int[] nums) {
    int len = nums.length;
    int left = 0, right = len - 1;
    if (nums[left] < nums[right]) {
        return 0;
    }
    while (left < right - 1) {
        int mid = left + (right - left) / 2;
        if (nums[mid] > nums[left]) {
            left = mid;
        } else {
            right = mid;
        }
    }
    return right;
}

// normal binary search, assume nums is ascending in [left, right]
private int binarySearch(int[] nums, int left, int right, int target) {
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) {
            return mid;
        } else if (nums[mid] < target) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    return nums[left] == target ? left : -1;
}

public int search(int[] nums, int target) {
    // cope with cases of length <= 1
    if (nums == null || nums.length == 0) {
        return -1;
    }
    if (nums.length == 1) {
        return target == nums[0] ? 0 : -1;
    }

    int len = nums.length;
    int pivot = findPivot(nums);
    // handle special case that there is only one interval
    if (pivot == 0) {
        return binarySearch(nums, 0, len - 1, target);
    }
    if (target >= nums[0]) {
        return binarySearch(nums, 0, pivot - 1, target);
    } else {
        return binarySearch(nums, pivot, len - 1, target);
    }
}