STL 源码剖析 算法 stl_algo.h -- nth_element

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nth_element

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描述：重新排序，使得[nth,last)内没有任何一个元素小于[first,nth)内的元素，

但对于[first,nth)和[nth,last)两个子区间内的元素次序则无任何保证。

思路：

1.以 median-of-3-partition 将整个序列分割为更小的左、右子序列

2.如果 nth 迭代器落于左序列，就再对左子序列进行分割，否则就再对右子序列进行分割

3.直到分割后的子序列长大于3，对最后这个待分割的子序列做 Insertion Sort



复杂度：O(n)

源码：

template <class RandomAccessIterator>
inline void nth_element(RandomAccessIterator first, RandomAccessIterator nth,
RandomAccessIterator last) {
__nth_element(first, nth, last, value_type(first));
}

template <class RandomAccessIterator, class T>
void __nth_element(RandomAccessIterator first, RandomAccessIterator nth,
RandomAccessIterator last, T*) {
while (last - first > 3) {
//采用 median-of-3-partition 。参数：(first,last,pivot)
//返回一个迭代器，指向分割后的右段第一个元素
RandomAccessIterator cut = __unguarded_partition
(first, last, T(__median(*first, *(first + (last - first)/2),
*(last - 1))));
if (cut <= nth) //如果  nth 落于右段，再对右段实施分割
first = cut;
else  //如果 nth 落于左段，对左段实施分割
last = cut;
}
__insertion_sort(first, last); //对分割后的子序列做 Insertion Sort
}

template <class RandomAccessIterator, class T>
RandomAccessIterator __unguarded_partition(RandomAccessIterator first,
RandomAccessIterator last,
T pivot) {
while (true) {
while (*first < pivot) ++first;
--last;
while (pivot < *last) --last;
if (!(first < last)) return first;
iter_swap(first, last);
++first;
}
}

示例：

int A[] = {7, 2, 6, 11, 9, 3, 12, 10, 8, 4, 1, 5};
const int N = sizeof(A) / sizeof(int);

nth_element(A, A + 6, A + N);
copy(A, A + N, ostream_iterator<int>(cout, " "));
// The printed result is "5 2 6 1 4 3 7 8 9 10 11 12".