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]. http://leetcode.com/onlinejudge#question_34

leetcode_search_for_a_range.cpp

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        int left = 0, right = n - 1;
        int ans1 = n;
        while (left <= right) {
            int mid = (left + right) >> 1;
            if (A[mid] == target) {
                ans1 = min(ans1, mid);
                right = mid - 1;
            }
            else if (A[mid] > target)
                right = mid - 1;
            else
                left = mid + 1;
        }
        ans1 = ans1 == n ? -1 : ans1;
        int ans2 = -1;
        left = 0, right = n - 1;
        while (left <= right) {
            int mid = (left + right) >> 1;
            if (A[mid] == target) {
                ans2 = max(ans2, mid);
                left = mid + 1;
            }
            else if (A[mid] > target)
                right = mid - 1;
            else
                left = mid + 1;
        }
        vector<int> ans;
        ans.push_back(ans1);
        ans.push_back(ans2);
        return ans;
    }
};