leetcode: Search for a Range

http://oj.leetcode.com/problems/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].

思路：

先用二分法找到target出现的位置，然后不停的用二分法确定左右区间。

class Solution {
public:
    int binarySearch(int A[], int start, int end, int target) {
        while (start <= end) {
            int mid = (start + end) / 2;
            
            if (A[mid] == target) {
                return mid;
            }
            else if (A[mid] < target) {
                start = mid + 1;
            }
            else {
                end = mid - 1;
            }
        }
        
        return -1;
    }

    
    vector<int> searchRange(int A[], int n, int target) {
        vector<int> result(2, -1);
        int pos = binarySearch(A, 0, n - 1, target);
        
        if (-1 == pos) {
            return result;
        }
        
        int tmp_pos, start = pos, end = pos;
        
        tmp_pos = pos - 1;
        
        while (tmp_pos >= 0) {
            tmp_pos = binarySearch(A, 0, tmp_pos, target);
            
            if (-1 == tmp_pos) {
                break;
            }
            else {
                start = tmp_pos;
                --tmp_pos;
            }
        }
        
        tmp_pos = pos + 1;
        
        while (tmp_pos < n) {
            tmp_pos = binarySearch(A, tmp_pos, n - 1, target);
            
            if (-1 == tmp_pos) {
                break;
            }
            else {
                end = tmp_pos;
                ++tmp_pos;
            }
        }
        
        result[0] = start;
        result[1] = end;
        
        return result;
    }
};