28.earch in Rotated Sorted Array(排序旋转数组中查找)
Level:
Medium
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
思路分析:
有序旋转数组,设置两个指针,left和right,分别指向数组的左端和右端,求数组的mid,如果mid的值大于left的值,那么旋转点在mid后面,如果小于left,则证明旋转点在mid前面。
如果旋点在mid的后面,并且target的值大于left处的值,小于mid处的值,那么接下来就可以在left到mid之间进行二分查找target,否则对mid+1到right这部分数组进行递归操作。
如果旋点在mid的前面,并且target的值大于mid处的值,小于right处的值,那么接下来就可以在mid到right之间进行二分查找target,否则对left到mid-1这部分数组进行递归操作。
代码:
public class Solution{
public int search(int []nums,int target){
if(nums==null||nums.length==0)
return -1;
int res=find(nums,0,nums.length-1,target);
return res;
}
public int find(int []nums,int left,int right,int target){
if(nums[left]==target)
return left;
if(nums[right]==target)
return right;
int mid=(left+right)/2;
if(nums[mid]==target)
return mid;
if(left>right)
return -1;
if(nums[mid]>nums[left]){ //证明旋转点在mid后面
if(target>nums[left]&&target<nums[mid]){
return binarySearch(nums,left,mid-1,target);
}else{
return find(nums,mid+1,right,target);
}
}
if(nums[mid]<nums[left]){ //证明旋转点在mid的前面
if(target>nums[mid]&&target<nums[right]){
return binarySearch(nums,mid+1,right,target);
}else{
return find(nums,left,mid-1,target);
}
}
return -1;
}
public int binarySearch(int []nums,int left,int right,int target){
if(left<=right){
int mid=(left+right)/2;
if(nums[mid]==target)
return mid;
if(target>nums[mid]){
return binarySearch(nums,mid+1,right,target);
}
if(target<nums[mid]){
return binarySearch(nums,left,mid-1,target);
}
}
return -1;
}
}