LeetCode33 搜索旋转数组

题目

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 

方法

二分查找法

二分查找，找到中点mid，判断左边递增还是右边递增，然后判断target是否在递增序列里，如果在则取递增一半，否则取另一半

• 时间复杂度：O(logn)
• 空间复杂度：O(1)

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