LC 33. Search in Rotated Sorted Array
问题描述
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
参考答案
class Solution { public: int search(vector<int>& nums, int target) { if(nums.size() == 0) return -1; int l = 0; int r = nums.size()-1; while(l<r){ int mid = (l+r)>>1; if(nums[l]<=nums[mid]){ if(target<=nums[mid] && target >= nums[l]) r = mid; else l = mid +1; }else if(nums[mid]<nums[r]){ if(target>nums[mid] && target <= nums[r]) l = mid +1; else r = mid; } } if(target == nums[l]) return r; return -1; } };
答案描述
由于不知道序列是否有序,因此在进行二分搜索的时候,需要先确定在什么区间:
1. 低位< 中位:意味着从 低位 到 中位,一定是有序的。如果循环的连接点,没有过中位,那么中位海拔更低,低位海拔更高,这个时候,有序的数列在,
2. 中位<高位:之间,这一段里面就是有序的,因此可以在这个区间内进行搜索。
宗旨,时刻保持有序。