public class Permutation {
    public static void main(String[] args) {
        Character[] ps = { 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L'};

        long timer = System.currentTimeMillis();
        int total = 0;
        total = fastPermutation(ps, 0);
        System.out.println("Got " + total + "\tspend: " + (System.currentTimeMillis() - timer) + "ms");
    }

    public static int fastPermutation(Character[] ps, int pos) {
        int cnt = 0;
        if (pos < ps.length - 1) {
            fastPermutation(ps, pos + 1);
            for (int i = pos + 1; i < ps.length; i++) {
                swap(ps, pos, i);
                cnt += fastPermutation(ps, pos + 1);
                swap(ps, pos, i);
            }
        } else {
            cnt++;
            //System.out.println(Arrays.toString(ps)); // 打开输出的话，会比较浪费时间。
        }
        return cnt;
    }

    private static void swap(Character[] ps, int pa, int pb) {
        Character tmp = ps[pa];
        ps[pa] = ps[pb];
        ps[pb] = tmp;

    }

------解决方案--------------------
写了个深搜。
暴力出所有情况。
但是，这是最不效率的方法。

应该有很高效的算法。


Java code


public class Rank {
    private char[] rank;
    private boolean[] vist;
    private char[] c;
    private int size;
    
    public Rank(char[] c){
        rank = new char[c.length];
        vist = new boolean[c.length];
        this.c = c;
        size = c.length;
    }
    public void showAllRank() {
        for (int i = 0; i < size; i++) {
            vist[i] = false;
        }
        dfs(0);
    }
    public void dfs(int level) {
        if (level == size) {
            for (int i = 0; i < size; i++) {
                System.out.print(rank[i]);
            }
            System.out.println();
        } else {
            for (int i = 0; i < size; i++) {
                if ( !vist[i]) {
                    vist[i] = true;
                    rank[level] = c[i];
                    dfs(level+1);
                    vist[i] = false; //回溯
                }
            }
        }
    }
    
}