LeetCode | 34. Search for a Range-优快云博客

本文介绍了一种在有序数组中快速查找指定目标值起始与结束位置的方法。通过两次二分查找，算法可在O(logn)时间内确定目标元素的范围下标。文章提供了完整的C++实现代码。

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;
}