递归问题（二）

在前文《递归问题一》中简单介绍了几个典型的递归问题。前几天看到一个人在论坛里问，如何实现一个字符串数组的全排列问题，底下人都说可以用递归方法实现，我想了会，没想出来，不过有人贴出了他的代码，我这里借用一下：

public class AllSort{ public static void main(String[] args) { char buf[]={'a','b','c'}; perm(buf,0,buf.length-1); } public static void perm(char[] buf,int start,int end){ if(start==end){//当只要求对数组中一个字母进行全排列时，只要就按该数组输出即可 for(int i=0;i<=end;i++){ System.out.print(buf[i]); } System.out.println(); } else{//多个字母全排列 for(int i=start;i<=end;i++){ char temp=buf[start];//交换数组第一个元素与后续的元素 buf[start]=buf[i]; buf[i]=temp; perm(buf,start+1,end);//后续元素递归全排列 temp=buf[start];//将交换后的数组还原 buf[start]=buf[i]; buf[i]=temp; } } } }

输出：

abc acb bac bca cba cab

还有一个比较著名的递归问题，就是n皇后问题。就是在一个n*n的国际象棋的方格中，怎么样才能将n个皇后摆在棋盘上共存。我们知道皇后是能横着，竖着和对角线这样三条路子。以8皇后为例讨论这个问题，那么为了和Java中的数组第一个索引0更好的结合起来，我们定义8*8的棋盘中每个方格的坐标为(0, 0)，(0, 1)...假定两个皇后(i, j)，(k, l)，那么这两个皇后的坐标一定得满足：(i != k) && (j != l) &&(|i-k| != |j-l|)。我们以一个长度为8的数组表示皇后的坐标，例如(0, 1, 2, ... , 7)，这就意味着第一个皇后摆在(0, 0)上，第二个皇后在(1, 1)上，依次类推。这样就能保证每个皇后的纵坐标是不一样的。下面是具体的实现代码：

class NQueen{ private int noOfSolutions; private int noOfQueens; private int[] queensInRow; public NQueen (){ this.noOfQueens = 8; this.queensInRow = new int[8]; this.noOfSolutions = 0; } public NQueen (int queens){ this.noOfQueens = queens; this.queensInRow = new int[noOfQueens]; this.noOfSolutions = 0; } public boolean canPlaceQueen (int k, int i){ for (int j=0; j<k; j++){ if ((queensInRow[j]==i) || (Math.abs(queensInRow[j]-i)== Math.abs(j-k))){ return false; } } return true; } public void queensConfiguration (int k){ for (int i=0; i<noOfQueens; i++){ if (canPlaceQueen(k, i)){ queensInRow[k] = i; if (k == noOfQueens-1){ printConfiguration(); } else{ queensConfiguration(k+1); } } } } public void printConfiguration (){ noOfSolutions ++; System.out.print("("); for (int i=0; i<noOfQueens-1; i++){ System.out.print(queensInRow[i] +", "); } System.out.println(queensInRow[noOfQueens-1] +")"); } public int solutionsCount(){ return noOfSolutions; } } public class TestNQueen { public static void main (String args[]){ NQueen queen = new NQueen(); queen.queensConfiguration(0); } }

输出：

(0, 4, 7, 5, 2, 6, 1, 3) (0, 5, 7, 2, 6, 3, 1, 4) (0, 6, 3, 5, 7, 1, 4, 2) (0, 6, 4, 7, 1, 3, 5, 2) (1, 3, 5, 7, 2, 0, 6, 4) (1, 4, 6, 0, 2, 7, 5, 3) (1, 4, 6, 3, 0, 7, 5, 2) (1, 5, 0, 6, 3, 7, 2, 4) (1, 5, 7, 2, 0, 3, 6, 4) (1, 6, 2, 5, 7, 4, 0, 3) (1, 6, 4, 7, 0, 3, 5, 2) (1, 7, 5, 0, 2, 4, 6, 3) (2, 0, 6, 4, 7, 1, 3, 5) (2, 4, 1, 7, 0, 6, 3, 5) (2, 4, 1, 7, 5, 3, 6, 0) (2, 4, 6, 0, 3, 1, 7, 5) (2, 4, 7, 3, 0, 6, 1, 5) (2, 5, 1, 4, 7, 0, 6, 3) (2, 5, 1, 6, 0, 3, 7, 4) (2, 5, 1, 6, 4, 0, 7, 3) (2, 5, 3, 0, 7, 4, 6, 1) (2, 5, 3, 1, 7, 4, 6, 0) (2, 5, 7, 0, 3, 6, 4, 1) (2, 5, 7, 0, 4, 6, 1, 3) (2, 5, 7, 1, 3, 0, 6, 4) (2, 6, 1, 7, 4, 0, 3, 5) (2, 6, 1, 7, 5, 3, 0, 4) (2, 7, 3, 6, 0, 5, 1, 4) (3, 0, 4, 7, 1, 6, 2, 5) (3, 0, 4, 7, 5, 2, 6, 1) (3, 1, 4, 7, 5, 0, 2, 6) (3, 1, 6, 2, 5, 7, 0, 4) (3, 1, 6, 2, 5, 7, 4, 0) (3, 1, 6, 4, 0, 7, 5, 2) (3, 1, 7, 4, 6, 0, 2, 5) (3, 1, 7, 5, 0, 2, 4, 6) (3, 5, 0, 4, 1, 7, 2, 6) (3, 5, 7, 1, 6, 0, 2, 4) (3, 5, 7, 2, 0, 6, 4, 1) (3, 6, 0, 7, 4, 1, 5, 2) (3, 6, 2, 7, 1, 4, 0, 5) (3, 6, 4, 1, 5, 0, 2, 7) (3, 6, 4, 2, 0, 5, 7, 1) (3, 7, 0, 2, 5, 1, 6, 4) (3, 7, 0, 4, 6, 1, 5, 2) (3, 7, 4, 2, 0, 6, 1, 5) (4, 0, 3, 5, 7, 1, 6, 2) (4, 0, 7, 3, 1, 6, 2, 5) (4, 0, 7, 5, 2, 6, 1, 3) (4, 1, 3, 5, 7, 2, 0, 6) (4, 1, 3, 6, 2, 7, 5, 0) (4, 1, 5, 0, 6, 3, 7, 2) (4, 1, 7, 0, 3, 6, 2, 5) (4, 2, 0, 5, 7, 1, 3, 6) (4, 2, 0, 6, 1, 7, 5, 3) (4, 2, 7, 3, 6, 0, 5, 1) (4, 6, 0, 2, 7, 5, 3, 1) (4, 6, 0, 3, 1, 7, 5, 2) (4, 6, 1, 3, 7, 0, 2, 5) (4, 6, 1, 5, 2, 0, 3, 7) (4, 6, 1, 5, 2, 0, 7, 3) (4, 6, 3, 0, 2, 7, 5, 1) (4, 7, 3, 0, 2, 5, 1, 6) (4, 7, 3, 0, 6, 1, 5, 2) (5, 0, 4, 1, 7, 2, 6, 3) (5, 1, 6, 0, 2, 4, 7, 3) (5, 1, 6, 0, 3, 7, 4, 2) (5, 2, 0, 6, 4, 7, 1, 3) (5, 2, 0, 7, 3, 1, 6, 4) (5, 2, 0, 7, 4, 1, 3, 6) (5, 2, 4, 6, 0, 3, 1, 7) (5, 2, 4, 7, 0, 3, 1, 6) (5, 2, 6, 1, 3, 7, 0, 4) (5, 2, 6, 1, 7, 4, 0, 3) (5, 2, 6, 3, 0, 7, 1, 4) (5, 3, 0, 4, 7, 1, 6, 2) (5, 3, 1, 7, 4, 6, 0, 2) (5, 3, 6, 0, 2, 4, 1, 7) (5, 3, 6, 0, 7, 1, 4, 2) (5, 7, 1, 3, 0, 6, 4, 2) (6, 0, 2, 7, 5, 3, 1, 4) (6, 1, 3, 0, 7, 4, 2, 5) (6, 1, 5, 2, 0, 3, 7, 4) (6, 2, 0, 5, 7, 4, 1, 3) (6, 2, 7, 1, 4, 0, 5, 3) (6, 3, 1, 4, 7, 0, 2, 5) (6, 3, 1, 7, 5, 0, 2, 4) (6, 4, 2, 0, 5, 7, 1, 3) (7, 1, 3, 0, 6, 4, 2, 5) (7, 1, 4, 2, 0, 6, 3, 5) (7, 2, 0, 5, 1, 4, 6, 3) (7, 3, 0, 2, 5, 1, 6, 4)

好了，希望这两篇blog能帮助你了解递归的思维方式，欢迎拍砖！