34. Search for a Range
来源:互联网 发布:东非解放军 知乎 编辑:程序博客网 时间:2024/03/29 14:19
Given a sorted array of integers, find the starting and ending position of a given target value.
Your algorithm’s runtime complexity must be in the order of O(log n).
If the target is not found in the array, return [-1, -1].
For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].
这道题要算法复杂度是O(log N),第一时间想到的就是二分法。
第一种方法:貌似不是二分法,但是也蛮快的。
class Solution {public: vector<int> searchRange(vector<int>& nums, int target) { vector<int>temp; int l=0,r=nums.size()-1; while(l<=r) { if(nums[l]==nums[r]&&nums[l]==target) { temp.push_back(l); temp.push_back(r); return temp; } f(target<nums[r]) r--; else if(target>nums[l]) l++; } temp.push_back(-1); temp.push_back(-1); return temp; }};
第二种方法:左右反复调用二分法,这样反而速度比较慢
class Solution {public: vector<int> searchRange(vector<int>& nums, int target) { vector<int>temp; int l=0,r=nums.size()-1; int left=-1; while(Binsearch(nums,target,l,r)!=-1) { left=Binsearch(nums,target,l,r); r=left-1; } temp.push_back(left); l=0,r=nums.size()-1; int right=-1; while(Binsearch(nums,target,l,r)!=-1) { right=Binsearch(nums,target,l,r); l=right+1; } temp.push_back(right); return temp; } int Binsearch(vector<int>& nums,int target,int left,int right) { int mid; while(left<=right) { mid=(left+right)/2; if(nums[mid]==target) return mid; else if(target>nums[mid]) left=mid+1; else right=mid-1; } return -1; }};
第三种方法只进行一次二分调用找到target中心,再向两边扩散
class Solution {public: vector<int> searchRange(vector<int>& nums, int target) { vector<int>temp; int l=0,r=nums.size()-1; int index=Binsearch(nums,target,l,r); if(index==-1) { temp.push_back(-1); temp.push_back(-1); } else { int left=index; while(left>0&&nums[left-1]==target) left--; int right=index; while(right<nums.size()-1&&nums[right+1]==target) right++; temp.push_back(left); temp.push_back(right); } return temp; } int Binsearch(vector<int>& nums,int target,int left,int right) { int mid; while(left<=right) { mid=(left+right)/2; if(nums[mid]==target) return mid; else if(target>nums[mid]) left=mid+1; else right=mid-1; } return -1; }};
0 0
- 34. Search for a Range
- 34. Search for a Range
- 34. Search for a Range
- 34. Search for a Range
- 34. Search for a Range
- 34. Search for a Range
- 34. Search for a Range
- 34. Search for a Range
- 34. Search for a Range
- 34. Search for a Range
- 34. Search for a Range
- 34. Search for a Range
- 34. Search for a Range
- 34. Search for a Range
- 34. Search for a Range
- 34. Search for a Range
- 34. Search for a Range
- 34. Search for a Range
- mysql学习之正则表达式查询
- REST软件架构
- GWT开发效率困境与解决之道
- APNS部署教程4(服务器/Provider开发)
- 解决overridePendingTransition失效的问题
- 34. Search for a Range
- Redis 数据结构使用场景
- Oauth协议
- 使用自定义证书并忽略验证的HTTPS连接Post请求方式的封装
- PyMongo使用入门(五)
- Leetcode 36 Valid Sudoku 数独的合法性判断
- Java 线程面试题
- foreach语句
- 携程基于Storm的实时大数据平台实践