33.Search-in-Rotated-Sorted-Array

33. Search in Rotated Sorted Array

题目地址

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

http://www.lintcode.com/problem/search-in-rotated-sorted-array/description

http://www.jiuzhang.com/solutions/search-in-rotated-sorted-array/

题目描述

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

代码

Approach #1 Binary Search

Complexity Analysis

• Time complexity: O(logN).

• Space complexity: O(1).

public class Solution {
    public int search(int[] A, int target) {
    if (A == null || A.length == 0) return -1;

    int start = 0;
    int end = A.length - 1;
    int mid;

    while (start + 1 < end) {
      mid = start + (end - start) / 2;
      if (A[mid] == target) {
        return mid;
      }

      if (A[start] < A[mid]) {
        if (A[start] <= target && target <= A[mid]) {
          end = mid;
        } else {
          start = mid;
        }
      } else {
        if (A[mid] <= target && target <= A[end]) {
          start = mid;
        } else {
          end = mid;
        }
      }
    }

    if (A[start] == target) {
      return start;
    }
    if (A[end] == target) {
      return end;
    }

    return -1; 
  }

}

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;
  }
}