

选择排序

Find the $i$-th smallest element (that is, the smallest element in $A_{i..n}$) each time, and then swap this element with the element at position $i$ of the array.

Time complexity is $O(n^2)$ .

Because of the operation of swapping two elements, the selection sort is an unstable sort.

Pseudocode:

$$\begin{array}{ll} 1 & \textbf{Input. } \text{An array } A \text{ consisting of }n\text{ elements.} \\ 2 & \textbf{Output. } A\text{ will be sorted in nondecreasing order.} \\ 3 & \textbf{Method. } \\ 4 & \textbf{for } i\gets 1\textbf{ to }n-1\\ 5 & \qquad ith\gets i\\ 6 & \qquad \textbf{for }j\gets i+1\textbf{ to }n\\ 7 & \qquad\qquad\textbf{if }A[j]<A[ith]\\ 8 & \qquad\qquad\qquad ith\gets j\\ 9 & \qquad \text{swap }A[i]\text{ and }A[ith]\\ \end{array}$$

C++ code:

void selection_sort(int* a, int n) {
  for (int i = 1; i < n; ++i) {
    int ith = i;
    for (int j = i + 1; j <= n; ++j) {
      if (a[j] < a[ith]) {
        ith = j;
      }
    }
    int t = a[i];
    a[i] = a[ith];
    a[ith] = t;
  }
}

---

build本页面最近更新：，更新历史
edit发现错误？想一起完善？ 在 GitHub 上编辑此页！
people本页面贡献者：OI-wiki
copyright本页面的全部内容在 CC BY-SA 4.0 和 SATA 协议之条款下提供，附加条款亦可能应用

评论