本文实例讲述了Java基于递归解决全排列问题算法。分享给大家供大家参考，具体如下：
排列问题
设R={r1,r2,...,rn}是要进行排列的n个元素，Ri=R-{ri}。集合x中元素的全排列记为Perm(X)。(ri)Perm(X)表示在全排列Perm(X)的每一个排列前加上前缀ri得到的排列。R的全排列可归纳如下：
当n=1时，Perm(R)=(r)，其中r是集合中唯一的元素；
当n>1时，Perm(R)由(r1)Perm(R1),(r2)Perm(R2),(r3)Perm(R3)。。。。(rn)Perm(Rn)构成。
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public class AllSort { public static void perm(int[] list, int k, int m) { if( k == m) { for (int i = 0; i <=m; i++) { System.out.print(list[i]); } System.out.println(); } else{ for(int i = k; i <= m; i++) { swap(list,k,i); perm(list, k+1 , m); swap(list,k,i); } } } public static void swap(int[] list, int a, int b) { int temp; temp = list[a]; list[a] = list[b]; list[b] = temp; } public static void main(String args[]) { int[] list = new int[5]; for(int i = 0; i < list.length; i++) { list[i] = i+1; } perm(list,0,list.length-1); }} |
运行结果：
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希望本文所述对大家java程序设计有所帮助。
原文链接：http://blog.csdn.net/u014755255/article/details/50183779
