LeetCode-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

使用二分查找，判断左右指针的位置，时间复杂度O(logn)

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