#Leetcode# 33. Search in Rotated Sorted Array
阿新 • • 發佈:2018-11-22
https://leetcode.com/problems/search-in-rotated-sorted-array/
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
程式碼:
class Solution { public: int search(vector<int>& nums, int target) { int n = nums.size(); map<int, int> mp; mp.clear(); for(int i = 0; i < n; i ++) mp[nums[i]] = i; sort(nums.begin(), nums.end()); int l = 0, r = n - 1; int mid; while(l <= r) { mid = (r - l) / 2 + l; if(nums[mid] == target) return mp[nums[mid]]; else if(nums[mid] > target) r = mid - 1; else l = mid + 1; } return -1; } };
驚險!0.93%
題解:原來是要找出來旋轉的位置 然鵝我只想到記下座標然後排序再二分 行吧行吧 哭唧唧 正解程式碼:
class Solution { public: int search(vector<int>& nums, int target) { int left = 0, right = nums.size() - 1; while (left <= right) { int mid = left + (right - left) / 2; if (nums[mid] == target) return mid; else if (nums[mid] < nums[right]) { if (nums[mid] < target && nums[right] >= target) left = mid + 1; else right = mid - 1; } else { if (nums[left] <= target && nums[mid] > target) right = mid - 1; else left = mid + 1; } } return -1; } };