Question:
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.
Tips:
给定一个数组,该数组是由一个有序的数组经过旋转(即将前面一段数字接到整个数组之后)得到的。判断target是否存在于该数组之中。
数组中不存在重复数字,如果target存在,返回他的index不存在则返回-1.
思路:
查找某数字是否存在,可以使用二分查找,但是由于该数组不是正常的有序,不能使用正常的BS。
设置两个指针,low high分别指向数组的首尾,mid=(high+low)/2;旋转后的数组用mid分为俩半,必然有一半是有序的,先判断target是不是在有序的一半内,如果在直接进行二分搜索,如果不在那么就在另外一部分里面。
代码:
public int search(int[] nums, int target) { if (nums == null || nums.length == 0) return -1; int len = nums.length ; int low = 0; int high = len - 1; while (low <= high) { int mid = low + (high - low) / 2; if (nums[mid] == target) return mid; if (nums[low] <= nums[mid]) { if(target<nums[mid] && target>=nums[low]){ high=mid-1; }else{ low=mid+1; } }else{ if(target>nums[mid] && target<=nums[high]){ low=mid+1; }else high=mid-1; } } return -1; }
原文地址:https://www.cnblogs.com/yumiaomiao/p/8530313.html
时间: 2024-10-29 05:18:48