LeetCode - Search in Rotated Sorted Array

Problem description

description

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

Analysis

Basic idea it to use binary search, and judge which part should go next.

Here’s the code:

public class SearchinRotatedSortedArray {
  public int search(int[] nums, int target) {
    return binarySearch(nums, target, 0, nums.length - 1);
  }

  int binarySearch(int[] nums, int target, int left, int right) {
    if (left < 0 || right < 0 || left >= nums.length || right >= nums.length) {
      return -1;
    }

    int mid = left + (right - left) / 2;
    if (nums[mid] == target) {
      return mid;
    } else if (nums[left] == target) {
      return left;
    } else if (nums[right] == target) {
      return right;
    } else {
      if (nums[left] < nums[right]) {
        if (target < nums[mid]) {
          return binarySearch(nums, target, left + 1, mid - 1);
        } else {
          return binarySearch(nums, target, mid + 1, right - 1);
        }
      } else {
        if (nums[mid] > nums[left]) {
          if (target > nums[mid] || target < nums[right]) {
            return binarySearch(nums, target, mid + 1, right - 1);
          } else {
            return binarySearch(nums, target, left + 1, mid - 1);
          }
        } else {
          if (target > nums[right] || target < nums[mid]) {
            return binarySearch(nums, target, left + 1, mid - 1);
          } else {
            return binarySearch(nums, target, mid + 1, right - 1);
          }
        }
      }
    }
  }
}

However, this is a ugly code, we only have to consider the increasing part and use else to go to the other disordered part.

Here’s the sample code from LeetCode:

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

What to improve

• should use else in a wise way