34. Search for a Range

题目

Given an array of integers sorted in ascending order, 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].

翻译

给定按升序排序的整数数组，找到给定目标值的起始和终止位置。

您的算法的运行时复杂度必须是O（log n）的顺序。

如果在数组中找不到目标，返回[-1, -1]。

例如，
给定[5, 7, 7, 8, 8, 10]和目标值8，
返回[3, 4]。

分析

采用二分法，因为要求target的左极限与右极限，所以要用二分法查找两次

时间复杂度O(logn)

递归法：

class Solution {public:    int Binaryup(vector<int>& nums,int target,int begin,int end){        if(begin>end) return end;        int mid=begin+(end-begin)/2;        if(nums[mid]>target)  return Binaryup(nums,target,begin,mid-1);        else return Binaryup(nums,target,mid+1,end);    }        int Binarylow(vector<int>& nums,int target,int begin,int end){        if(begin>end)  return begin;        int mid=begin+(end-begin)/2;        if(nums[mid]<target) return Binarylow(nums,target,mid+1,end);        else return Binarylow(nums,target,begin,mid-1);    }    vector<int> searchRange(vector<int>& nums, int target) {        vector<int> result(2,-1);        int h=Binaryup(nums,target,0,nums.size()-1);        int l=Binarylow(nums,target,0,nums.size()-1);        if(h>=l){            result[0]=l;            result[1]=h;            return result;        }        return result;    }};

while循环：

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        vector<int> res(2,-1);        if(nums.empty())            return res;        int l=0,r=nums.size()-1;        while(l<r){            int m=l+(r-l)/2;            if(nums[m]==target) r=m;            else if(nums[m]<target) l=m+1;            else r=m-1;        }        if(nums[l]==target)            res[0]=l;        else             return res;        r=nums.size()-1;        while(l<r){            int m=l+(r-l)/2+1;            if(nums[m]==target) l=m;            else if(nums[m]<target) l=m+1;            else r=m-1;        }        res[1]=l;        return res;    }};