[leedcode 33] Search in Rotated Sorted Array

Suppose a sorted array 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.

public class Solution {
    public int search(int[] nums, int target) {
        //二分查找法的变形，利用数组特点，每次切掉一半
        //注意比较时等号的使用，以及比较的条件if(target>=nums[left]&&target<nums[mid])
        //本题还可以利用递归进行解决,递归貌似没有利用数组的特点，所有二分法都通用
       /* if(nums.length<1) return -1;
        int left=0;
        int right=nums.length-1;
        while(left<=right){
            if(left==right) return nums[left]==target?left:-1;
            int mid=(left+right)/2;
            if(nums[mid]==target){
                return mid;
            }
            if(nums[mid]>=nums[left]){//注意等号
                if(target>=nums[left]&&target<nums[mid])//注意等号
                  right=mid-1;
                 else{
                  left=mid+1;
                    
                 }
                
            }else{
                if(target>nums[mid]&&target<=nums[right]){//注意等号
                    left=mid+1;
                }else{
                    right=mid-1;
                }
            }
            
            
        }
        return -1;*/
        //if(nums.length<1)return -1;
        return find(nums,target,0,nums.length-1);
        
    }
    public int find(int []num,int target,int start,int end){
        if(start>end) return -1;
        int mid=(start+end)/2;
        if(num[mid]==target) return mid;
        int res=find(num,target,start,mid-1);
        if(res==-1) //若果左侧没有找到，进行右侧的查找
            res=find(num,target,mid+1,end);
        return res;
        
    }
    
    
}