[LeetCode] Search for a Range

Given a sorted array of integers, 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].

二分找最左端，再二分找最右端。

class Solution {
public:
    int findPos(int a[], int beg, int end, int key, bool findLeft)
    {
        if (beg > end)
            return -1;
            
        int mid = (beg + end) / 2;
        
        if (a[mid] == key)
        {
            int pos = findLeft ? findPos(a, beg, mid - 1, key, findLeft) : findPos(a, mid + 1, end, key, findLeft);
            return pos == -1 ? mid : pos;
        }
        else if (a[mid] < key)
            return findPos(a, mid + 1, end, key, findLeft);
        else
            return findPos(a, beg, mid - 1, key, findLeft);       
    }
    
    vector<int> searchRange(int A[], int n, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int leftPos = findPos(A, 0, n - 1, target, true);
        int rightPos = findPos(A, 0, n - 1, target, false);
        
        vector<int> ret;
        
        ret.push_back(leftPos);
        ret.push_back(rightPos);
        return ret;
    }
};