33. Search in Rotated Sorted Array

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Your algorithm's runtime complexity must be in the order of O(log n).

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

Solution

(1) Java

class Solution {
    public int search(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        int end = nums.length-1;
        int start = 0;
        while (start+1 < end) {
            int mid = start + (end-start)/2;
            if (nums[mid] == target) {
                return mid;
            }
            if (nums[mid] > nums[start] && nums[mid] < nums[end]) {
                if (nums[mid] > target) {
                    end = mid;
                } else {
                    start = mid+1;
                }
            } else if (nums[mid] > nums[start]) {
                if (nums[mid] > target && nums[start] <= target) {
                    end = mid;
                } else {
                    start = mid+1;
                }
            } else {
                if (nums[mid] < target && nums[end] >= target) {
                    start = mid+1;
                } else {
                    end = mid;
                }
            }
        }
        if (nums[start] == target) {
            return start;
        } else if (nums[end] == target) {
            return end;
        }
        return -1;
    }
}

(2) Python

class Solution:
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        if not nums:
            return -1
        start, end = 0, len(nums)-1
        while start < end:
            mid = int(start + (end-start) / 2)
            if nums[mid] == target:
                return mid
            if nums[mid] >= nums[start]:
                if target >= nums[start] and target < nums[mid]:
                    end = mid
                else:
                    start = mid+1
            else:
                if target > nums[mid] and target <= nums[end]:
                    start = mid+1
                else:
                    end = mid-1
        if nums[start] == target:
            return start
        return -1

(3) Scala