【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].

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        vector<int> ret(2, -1);
        int start = 0, end = n - 1;
        int mid;
        bool found = false;
        while (start <= end && !found) {
            mid = (start + end) / 2;
            if (A[mid] == target) {
                found = true;
            } else if (A[mid] < target) {
                start = mid + 1;
            } else {
                end = mid - 1;
            }
        }
        if (found) {
            start = mid;
            while (start - 1 >= 0 && A[start - 1] == target) {
                --start;
            }
            end = mid;
            while (end + 1 < n && A[end + 1] == target) {
                ++end;
            }
            ret[0] = start;
            ret[1] = end;
        }
        return ret;
    }
};

二分查找。