[leetcode] Search for a Range

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].

 

分析：二分查找

class Solution
{
public:
  void Searching(vector<int> &ret, int A[], int s, int e, int target)
  {
    if(s > e)
    {
      ret.push_back(-1);
      ret.push_back(-1);
      return;
    }

    int mid = (s+e)/2;
    if(A[mid] < target)
      Searching(ret, A, mid+1, e, target);
    else if(A[mid] > target)
      Searching(ret, A, s, mid-1, target);
    else
    {
      int i = mid - 1;
      while(i >= s && A[i] == A[mid]) i--;
      ret.push_back(i + 1);

      i = mid + 1;
      while(i <= e && A[i] == A[mid]) i++;
      ret.push_back(i - 1);
    }
  }
  vector<int> searchRange(int A[], int n, int target)
  {
    vector<int> ret;
    if(target < A[0] || target > A[n-1])
    {
      ret.push_back(-1);
      ret.push_back(-1);
    }
    else
      Searching(ret, A, 0, n-1, target);

    return ret;
  }
};