标签：while vector algo 时间复杂度 exists ota inpu tor duplicate
算法描述：
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
解题思路：时间复杂度限制为O（log n）则首先想到用二分法求解。这道题的要点是先判断哪一部分是递增哪一部分是非递增的，或者都是递增的。然后再分段讨论。
int search(vector<int>& nums, int target) { int low = 0; int high = nums.size()-1; while(low <= high){ int mid = low + (high - low) /2; if(nums[mid]==target) return mid; if(nums[mid] > nums[high]){ if(nums[low] <= target && target < nums[mid]) high = mid-1; else low = mid+1; }else if(nums[mid] < nums[low]){ if(nums[mid] < target && target <= nums[high]) low = mid+1; else high = mid-1; } else { if(nums[mid] < target) low = mid +1; else high = mid -1; } } return -1; }
LeetCode-33-Search in Rotated Sorted Array
标签：while vector algo 时间复杂度 exists ota inpu tor duplicate
原文地址：https://www.cnblogs.com/nobodywang/p/10331966.html
