LeetCode | 34. Search for a Range

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].

题意：在O(log n )的时间内找到给定target元素在数组中的范围下标。
方法：用两次二分，找出第一个比它大的(mmax)和最后一个比它小的(mmin)。

#include <iostream>#include <vector>using namespace std;vector<int> searchRange(vector<int>& nums, int target){    //先找到最后一个比它小的，再找第一个比它大的    int len = nums.size();    int flag = -1;       //判断是否有target    int mmin = -1, mmax = len;    int left = 0, right = len-1;    vector<int> res;    while(left <= right)    {        int mid = (left+right)/2;        if(nums[mid] == target)        {            flag = mid; break;        }        else if(nums[mid] < target)            left = mid+1;        else            right = mid-1;    }    if(flag == -1)       //找不到target    {        mmin = -1; mmax = -1;        res.push_back(mmin);        res.push_back(mmax);        return res;    }    else    {        //cout<<flag<<endl;        //找出最后一个比它小的        left = 0; right = flag;        while(left <= right)        {            int mid = (left+right)/2;            if(nums[mid] < target)            {                if(mid+1 <= flag)                {                    if(nums[mid+1] >= target)                    {                        mmin = mid; break;                    }                    else                    {                        left = mid+1;                    }                }                else                    break;            }            else//比target大            {                right = mid-1;            }        }        //cout<<mmin<<endl;        //找出第一个比它大的        left = flag; right = len-1;        while(left <= right)        {            int mid = (left+right)/2;            if(nums[mid] > target)            {                if(mid-1 >= 0)                {                    if(nums[mid-1] <= target)                    {                        mmax = mid; break;                    }                    else                    {                        right = mid-1;                    }                }                else                    break;            }            else//不比target大            {                left = mid+1;            }        }    }    res.push_back(mmin+1);    res.push_back(mmax-1);    return res;}int main(){    vector<int> v;    int num, target;    while(cin>>num)    {        v.clear();        for(int i=0;i<num;i++)        {            int t;            cin>>t;            v.push_back(t);        }        cin>>target;        vector<int>res = searchRange(v,target);        for(int i=0;i<res.size();i++)        {            cout<<res[i]<<" ";        }        cout<<endl;    }    return 0;}