61. Search for a Range【medium】

61. Search for a Range【medium】

Given a sorted array of n integers, find the starting and ending position of a given target value.

If the target is not found in the array, return [-1, -1].

Have you met this question in a real interview? 

Yes

Example

Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

Challenge 

O(log n) time.

解法一：

class Solution {
public:
    /*
     * @param A: an integer sorted array
     * @param target: an integer to be inserted
     * @return: a list of length 2, [index1, index2]
     */
    vector<int> searchRange(vector<int> &A, int target) {
        if (A.empty()) {
            return vector<int>(2, -1);
        }
        
        vector<int> result;
        
        //find first
        int start = 0;
        int end = A.size() - 1;
        
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (A[mid] == target) {
                end = mid;
            }
            else if (A[mid] < target) {
                start = mid;
            }
            else if (A[mid] > target) {
                end = mid;
            }
        }
        
        if (A[start] == target) {
            result.push_back(start);
        }
        else if (A[end] == target) {
            result.push_back(end);
        }
        else {
            return vector<int>(2, -1);
        }
        
        //find last
        start = 0;
        end = A.size() - 1;
        
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (A[mid] == target) {
                start = mid;
            }
            else if (A[mid] < target) {
                start = mid;
            }
            else if (A[mid] > target) {
                end = mid;
            }
        }
        
        if (A[end] == target) {
            result.push_back(end);
        }
        else if (A[start] == target) {
            result.push_back(start);
        }
        
        return result;
    }
};