STL 源代码剖析 算法 stl_algo.h -- lower_bound

Posted 2020-06-29 Brenda

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lower_bound(应用于有序区间)

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描写叙述：二分查找，返回一个迭代器指向每个"不小于 value "的元素，
或 value 应该存在的位置
思路：
1.循环直到区间长度为 0 
2.假设 *middle < value，在后半段继续查找
3.假设 *middle >= value，在前半段继续查找 (等于的时候也会继续在前半段查找，所以能保证找到的是 lower bound)
源代码：

template <class ForwardIterator, class T>
inline ForwardIterator lower_bound(ForwardIterator first, ForwardIterator last,
                                   const T& value) {
  return __lower_bound(first, last, value, distance_type(first),
                       iterator_category(first));
}


// forward_iterator_tag 版本号
template <class ForwardIterator, class T, class Distance>
ForwardIterator __lower_bound(ForwardIterator first, ForwardIterator last,
                              const T& value, Distance*,
                              forward_iterator_tag) {
  Distance len = 0;
  distance(first, last, len);
  Distance half;
  ForwardIterator middle;


  while (len > 0) {
    half = len >> 1;
    middle = first;
    advance(middle, half); // 由于仅仅是 ForwardIterator，不能採用 middle = middle + half 的方式 
    if (*middle < value) {
      first = middle;
      ++first;
      len = len - half - 1;
    } // 由于 *middle >= value 时，会在前半段继续查找。所以终于找到的是 lower bound
    else
      len = half;
  }
  return first;
}


// random_access_iterator_tag 版本号
template <class RandomAccessIterator, class T, class Distance>
RandomAccessIterator __lower_bound(RandomAccessIterator first,
                                   RandomAccessIterator last, const T& value,
                                   Distance*, random_access_iterator_tag) {
  Distance len = last - first; // 整个区间长度
  Distance half;
  RandomAccessIterator middle;


  while (len > 0) {
    half = len >> 1; //除以2
    middle = first + half;
    if (*middle < value) { 
      first = middle + 1;
      len = len - half - 1; // -half-1 是由于前面那段有first指向的元素和half指向的区间 
    }
    else //为什么这种代码能保证找到的是 lower bound ？--> 由于小于等于都是到前面一段区间查找，所以最后找到的一定是 lower bound
      len = half;
  }
  return first;
}

演示样例：

int main()
{
  int A[] = { 1, 2, 3, 3, 3, 5, 8 };
  const int N = sizeof(A) / sizeof(int);


  for (int i = 1; i <= 10; ++i) {
    int* p = lower_bound(A, A + N, i);
    cout << "Searching for " << i << ".  ";
    cout << "Result: index = " << p - A << ", ";
    if (p != A + N)
      cout << "A[" << p - A << "] == " << *p << endl;
    else
      cout << "which is off-the-end." << endl;
  }
}
/*
The output is:
Searching for 1.  Result: index = 0, A[0] == 1
Searching for 2.  Result: index = 1, A[1] == 2
Searching for 3.  Result: index = 2, A[2] == 3
Searching for 4.  Result: index = 5, A[5] == 5
Searching for 5.  Result: index = 5, A[5] == 5
Searching for 6.  Result: index = 6, A[6] == 8
Searching for 7.  Result: index = 6, A[6] == 8
Searching for 8.  Result: index = 6, A[6] == 8
Searching for 9.  Result: index = 7, which is off-the-end.
Searching for 10.  Result: index = 7, which is off-the-end.
*/