34. Find First and Last Position of Element in Sorted Array

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]


输出某个数的区间

C++：

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int first = binarySearch(nums,target) ;
        int last = binarySearch(nums,target+1) ;
        if (first == nums.size() || nums[first] != target){
            return vector<int>(2,-1) ;
        }
        vector<int> res ;
        res.push_back(first) ;
        res.push_back(max(first,last-1)) ;
        return res ;
    }
    
    int binarySearch(vector<int> nums, int target) {
        int left = 0 ;
        int right = nums.size() ;
        while(left < right){
            int mid = left + (right - left) / 2 ;
            if (nums[mid] >= target){
                right = mid ;
            }else{
                left = mid + 1 ;
            }
        }
        return left ;
    }
};