【编程题】搜索旋转的排序数组(java实现)
题目来源
leetcode第33题
https://leetcode-cn.com/problems/search-in-rotated-sorted-array/submissions/
题目描述
假设按照升序排序的数组在预先未知的某个点上进行了旋转。
( 例如,数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。
搜索一个给定的目标值,如果数组中存在这个目标值,则返回它的索引,否则返回 -1 。
你可以假设数组中不存在重复的元素。
你的算法时间复杂度必须是 O(log n) 级别。
示例 1:
输入: nums = [4,5,6,7,0,1,2], target = 0
输出: 4
示例 2:
输入: nums = [4,5,6,7,0,1,2], target = 3
输出: -1
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/search-in-rotated-sorted-array
题目代码
leetcode代码
class Solution {
public int search(int[] nums, int target) {
if(nums==null)
return -1;
int left=0,right=nums.length-1;
while (left<right){
int mid=(left+right)>>1;
if(nums[mid]==target)
return mid;
else if(nums[0]<=nums[mid]&&(nums[mid]>=target&&nums[0]<=target)){
right=mid-1;
}
else if(nums[0]>nums[mid]&&nums[mid]>=target&&target<nums[0]){
right=mid-1;
}
else if(nums[0]>nums[mid]&&nums[mid]<target&&nums[0]<=target)
right=mid-1;
else
left=mid+1;
}
if(left==right&&nums[left]==target)
return left;
return -1;
}
}
包含main方法的IDEA代码
package practice;
public class Rotate {
public static void main(String[] args){
int[] arr={4,5,6,7,1,2,3};
System.out.println(findInRotate(arr,3));
}
static int findInRotate(int[] nums,int target){
if(nums==null)
return -1;
int left=0,right=nums.length-1;
while (left<right){
int mid=(left+right)>>1;
if(nums[mid]==target)
return mid;
else if(nums[0]<=nums[mid]&&(nums[mid]>=target&&nums[0]<=target)){
right=mid-1;
}
else if(nums[0]>nums[mid]&&nums[mid]>=target&&target<nums[0]){
right=mid-1;
}
else if(nums[0]>nums[mid]&&nums[mid]<target&&nums[0]<=target)
right=mid-1;
else
left=mid+1;
}
if(left==right&&nums[left]==target)
return left;
return -1;
}
}