[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].

二分法。一开始没完全理解题意,误以为返回两个相邻或相等的index的情况就可以了。考虑以下情况:

Given [5,7,7,8,8,8] Target Value: 8

return [3,5]


class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
    	vector<int> result;
		result.push_back(findPosition(A,0,n-1,target,true));
		result.push_back(findPosition(A,0,n-1,target,false));
		return result;
        
    }
	int findPosition (int A[], int begin, int end, int target, bool leftFlag)
	{
		if(begin>end) return -1;
		int middle = (begin + end)/2;
		if(A[middle] == target)
		{
			int position;
			if(leftFlag)
			{
				position = findPosition(A,begin,middle-1,target,leftFlag);
			}
			else
			{
				position = findPosition(A,middle+1,end,target,leftFlag);
			}
			return (position == -1)?middle:position;
		}
		else if(A[middle]<target)
		{
			return findPosition(A,middle+1,end,target,leftFlag);
		}
		else
		{
			return findPosition(A,begin,middle-1,target,leftFlag);
		}
	}

};


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