STL 源码剖析 算法 stl_algo.h -- search_n

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search_n

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描述：在序列[first, last) 所涵盖的区间中，查找"连续 count 个符合条件之元素"所形成的子序列，

并返回迭代器 last

思路：

1.首先找出 value 第一次出现点

2.该出现点的后面是否连续出现 count - 1 个 value

3.如果是，找到了，如果不是，在当前元素后的区间重新找 value 的出现点

图6-6k

template <class ForwardIterator, class Integer, class T>
ForwardIterator search_n(ForwardIterator first, ForwardIterator last,
                         Integer count, const T& value) {
  if (count <= 0)
    return first;
  else {
    first = find(first, last, value); // 首先找出 value 第一次出现点
    while (first != last) { // 这里的条件写成 last - first < n 是不是好些？
      Integer n = count - 1; // value 还应该出现 n 次
      ForwardIterator i = first;
      ++i;
      while (i != last && n != 0 && *i == value) {
        ++i;
        --n;
      }
      if (n == 0) // 找到了
        return first;
      else  // 没找到，重新从 i 开始找
        first = find(i, last, value);
    }
    return last;
  }
}

示例：

bool eq_nosign(int x, int y) { return abs(x) == abs(y); }
void lookup(int* first, int* last, size_t count, int val) {
  cout << "Searching for a sequence of "
       << count
       << " '" << val << "'"
       << (count != 1 ? "s: " : ":  ");
  int* result = search_n(first, last, count, val);
  if (result == last)
    cout << "Not found" << endl;
  else
    cout << "Index = " << result - first << endl;
}

void lookup_nosign(int* first, int* last, size_t count, int val) {
  cout << "Searching for a (sign-insensitive) sequence of "
       << count
       << " '" << val << "'"
       << (count != 1 ? "s: " : ":  ");
  int* result = search_n(first, last, count, val, eq_nosign);
  if (result == last)
    cout << "Not found" << endl;
  else
    cout << "Index = " << result - first << endl;
}

int main() {
  const int N = 10;
  int A[N] = {1, 2, 1, 1, 3, -3, 1, 1, 1, 1};

  lookup(A, A+N, 1, 4);
  lookup(A, A+N, 0, 4);
  lookup(A, A+N, 1, 1);
  lookup(A, A+N, 2, 1);
  lookup(A, A+N, 3, 1);
  lookup(A, A+N, 4, 1);

  lookup(A, A+N, 1, 3);
  lookup(A, A+N, 2, 3);
  lookup_nosign(A, A+N, 1, 3);
  lookup_nosign(A, A+N, 2, 3);
}
/*
The output is
Searching for a sequence of 1 '4':  Not found
Searching for a sequence of 0 '4's: Index = 0
Searching for a sequence of 1 '1':  Index = 0
Searching for a sequence of 2 '1's: Index = 2
Searching for a sequence of 3 '1's: Index = 6
Searching for a sequence of 4 '1's: Index = 6
Searching for a sequence of 1 '3':  Index = 4
Searching for a sequence of 2 '3's: Not found
Searching for a (sign-insensitive) sequence of 1 '3':  Index = 4
Searching for a (sign-insensitive) sequence of 2 '3's: Index = 4
*/

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