LeetCode : 33. Search in Rotated Sorted Array 旋转数组的二分搜索

试题
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
这一题与二分搜索的差别在于，存在一个区域纠正措施。
对于mid位置我们首先判断他是在左边递增序列中还是在右边递增序列中。
然后根据数组的首尾值进行一个区间的纠正。

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