描述：

Given an array of integers nums 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].

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

思路一：一次Binary Search

先使用二分查找找到和taget相等的起始位置的元素，然后在这个位置向两边扩散找到值相同的范围。

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> res(2,-1);
        if(nums.size() == 0) return res;
        int cau = dichotomy(nums,target);
        if(cau == -1) return res;
        else{
            int i = cau,j = cau;
            cout<<cau<<endl;
            while((i>=0 && nums[i] == target) || (j<=nums.size()-1 && nums[j] == target)){
                if(i>=0 && nums[i] == target){
                    res[0] = i;
                    cout<<i<<endl;
                    i--;
                }
                if(j<=nums.size()-1 && nums[j] == target){
                    res[1] = j;
                    cout<<j<<endl;
                    j++;
                }
            }
        }
        return res;
    }

    int dichotomy(vector<int>& nums, int target){
        int low = 0,int high = nums.size() - 1;
        while(low < high){
            int mid = (low + high + 1) / 2;
            if(nums[mid] == target) return mid;
            if(nums[mid] < target) low = mid + 1;
            else high = mid - 1;
        }
        return -1;
    }
};

思路二：两3次Binary Search

先根据上述的二分查找，找到起始位置的元素，然后从low开始继续使用一次二分查找找到target+1的位置，然后返回这个位置it - 1，从而找到起始和终止的范围。

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> res(2,-1);
        if(nums.size() == 0) return res;
        int low = dichotomy(nums,target,0);   //找起始元素
        cout<<low<<endl;
        if(low == nums.size() || nums[low] != target)  //可能出现到数组结尾还是target比low处元素大，low=nums.size
            return res;
        else{
            res[0] = low;
            res[1] = dichotomy(nums,target+1,low) - 1;   //找结尾元素
        }
        return res;
    }

    int dichotomy(vector<int>& nums, int target, int low){
        int high = nums.size(); //核心
        while(low < high){
            int mid = (low + high ) / 2;
            if(nums[mid] < target) low = mid + 1;
            else high = mid;
        }
        return low;
    }
};

原文地址：https://www.cnblogs.com/ygh1229/p/9726966.html