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深度优先遍历-解决排列组合问题
问题1：
假设袋子里有编号为1,2,…,m这m个球。现在每次从袋子中取一个球记下编号，放回袋中再取，取n次作为一组，枚举所有可能的情况。
分析：
每一次取都有m种可能的情况，因此一共有 m n m^n mn种情况。
这里我们取m = 3, n = 4，则有 3 4 3^4 34种不同的情况。
代码：
import java.util.Stack;public class Test {static int cnt = 0;static Stack<Integer> s = new Stack<Integer>();/*** 递归方法，当实际选取的小球数目与要求选取的小球数目相同时，跳出递归* @param minv - 小球编号的最小值* @param maxv - 小球编号的最大值* @param curnum - 当前已经确定的小球的个数* @param maxnum - 要选取的小球的数目*/public static void kase1(int minv,int maxv,int curnum, int maxnum){if(curnum == maxnum){cnt++;System.out.println(s);return;}for(int i = minv; i <= maxv; i++){s.push(i);kase1(minv, maxv, curnum+1, maxnum);s.pop();}}public static void main(String[] args){kase1(1, 3, 0, 4);System.out.println(cnt);}
}
输出：
[1, 1, 1, 1]
[1, 1, 1, 2]
[1, 1, 1, 3]
[1, 1, 2, 1]
[1, 1, 2, 2]
[1, 1, 2, 3]
[1, 1, 3, 1]
[1, 1, 3, 2]
[1, 1, 3, 3]
[1, 2, 1, 1]
[1, 2, 1, 2]
[1, 2, 1, 3]
[1, 2, 2, 1]
[1, 2, 2, 2]
[1, 2, 2, 3]
[1, 2, 3, 1]
[1, 2, 3, 2]
[1, 2, 3, 3]
[1, 3, 1, 1]
[1, 3, 1, 2]
[1, 3, 1, 3]
[1, 3, 2, 1]
[1, 3, 2, 2]
[1, 3, 2, 3]
[1, 3, 3, 1]
[1, 3, 3, 2]
[1, 3, 3, 3]
[2, 1, 1, 1]
[2, 1, 1, 2]
[2, 1, 1, 3]
[2, 1, 2, 1]
[2, 1, 2, 2]
[2, 1, 2, 3]
[2, 1, 3, 1]
[2, 1, 3, 2]
[2, 1, 3, 3]
[2, 2, 1, 1]
[2, 2, 1, 2]
[2, 2, 1, 3]
[2, 2, 2, 1]
[2, 2, 2, 2]
[2, 2, 2, 3]
[2, 2, 3, 1]
[2, 2, 3, 2]
[2, 2, 3, 3]
[2, 3, 1, 1]
[2, 3, 1, 2]
[2, 3, 1, 3]
[2, 3, 2, 1]
[2, 3, 2, 2]
[2, 3, 2, 3]
[2, 3, 3, 1]
[2, 3, 3, 2]
[2, 3, 3, 3]
[3, 1, 1, 1]
[3, 1, 1, 2]
[3, 1, 1, 3]
[3, 1, 2, 1]
[3, 1, 2, 2]
[3, 1, 2, 3]
[3, 1, 3, 1]
[3, 1, 3, 2]
[3, 1, 3, 3]
[3, 2, 1, 1]
[3, 2, 1, 2]
[3, 2, 1, 3]
[3, 2, 2, 1]
[3, 2, 2, 2]
[3, 2, 2, 3]
[3, 2, 3, 1]
[3, 2, 3, 2]
[3, 2, 3, 3]
[3, 3, 1, 1]
[3, 3, 1, 2]
[3, 3, 1, 3]
[3, 3, 2, 1]
[3, 3, 2, 2]
[3, 3, 2, 3]
[3, 3, 3, 1]
[3, 3, 3, 2]
[3, 3, 3, 3]
81
问题2：
假设袋子里有编号为1,2,…,m这m个球。先后从袋子中取出n个球，依次记录编号，枚举所有可能的情况。
分析：
这是排列问题，如果取出的球顺序不同，也是算不同的情况。因此应该有 m ∗ ( m − 1 ) ∗ ( m − 2 ) ∗ . . . ∗ ( m − n + 1 ) m*(m-1)*(m-2)*...*(m-n+1) m∗(m−1)∗(m−2)∗...∗(m−n+1)种情况，即 A m n = m ! ( m − n ) ! A_m^n=\frac{m!}{(m-n)!} Amn=(m−n)!m!种
这里取m = 5, n = 3。则有5*4*3种。
和问题1相比，唯一的区别是排列中不可以有重复。因此开了used数组用以标记是否已经访问。
代码：
import java.util.Stack;public class Test {static int cnt = 0;static Stack<Integer> s = new Stack<Integer>();static boolean[] used = new boolean[10000];/*** 递归方法，当实际选取的小球数目与要求选取的小球数目相同时，跳出递归* @param minv - 小球编号的最小值* @param maxv - 小球编号的最大值* @param curnum - 当前已经确定的小球的个数* @param maxnum - 要选取的小球的数目*/public static void kase2(int minv,int maxv,int curnum, int maxnum){if(curnum == maxnum){cnt++;System.out.println(s);return;}for(int i = minv; i <= maxv; i++){if(!used[i]){ //判断是否已经取过s.push(i);used[i] = true;kase2(minv, maxv, curnum+1, maxnum);s.pop();used[i] = false;}}}public static void main(String[] args){kase2(1, 5, 0, 3);System.out.println(cnt);}
}
输出：
[1, 2, 3]
[1, 2, 4]
[1, 2, 5]
[1, 3, 2]
[1, 3, 4]
[1, 3, 5]
[1, 4, 2]
[1, 4, 3]
[1, 4, 5]
[1, 5, 2]
[1, 5, 3]
[1, 5, 4]
[2, 1, 3]
[2, 1, 4]
[2, 1, 5]
[2, 3, 1]
[2, 3, 4]
[2, 3, 5]
[2, 4, 1]
[2, 4, 3]
[2, 4, 5]
[2, 5, 1]
[2, 5, 3]
[2, 5, 4]
[3, 1, 2]
[3, 1, 4]
[3, 1, 5]
[3, 2, 1]
[3, 2, 4]
[3, 2, 5]
[3, 4, 1]
[3, 4, 2]
[3, 4, 5]
[3, 5, 1]
[3, 5, 2]
[3, 5, 4]
[4, 1, 2]
[4, 1, 3]
[4, 1, 5]
[4, 2, 1]
[4, 2, 3]
[4, 2, 5]
[4, 3, 1]
[4, 3, 2]
[4, 3, 5]
[4, 5, 1]
[4, 5, 2]
[4, 5, 3]
[5, 1, 2]
[5, 1, 3]
[5, 1, 4]
[5, 2, 1]
[5, 2, 3]
[5, 2, 4]
[5, 3, 1]
[5, 3, 2]
[5, 3, 4]
[5, 4, 1]
[5, 4, 2]
[5, 4, 3]
60
问题3：
从m个球里（编号为1,2,3…,m）一次取n个球，其中m>n，记录取出球的编号，枚举所有的可能性。
分析：
这是组合问题。应该有 ( m n ) = m ! n ! ( m − n ) ! \binom{m}{n}=\frac{m!}{n!(m-n)!} (nm)=n!(m−n)!m!种可能性。
这里，如果取m = 8, n = 4. 则有 ( 8 4 ) = 8 ! 4 ! ( 8 − 4 ) ! = 8 × 7 × 6 × 5 4 × 3 × 2 × 1 = 70 \binom{8}{4}=\frac{8!}{4!(8-4)!}=\frac{8\times7\times6\times5}{4\times3\times2\times1}=70 (48)=4!(8−4)!8!=4×3×2×18×7×6×5=70种可能。
代码：
import java.util.Stack;public class Test {static int cnt = 0;static Stack<Integer> s = new Stack<Integer>();/*** 递归方法，当前已抽取的小球个数与要求抽取小球个数相同时，退出递归* @param curnum - 当前已经抓取的小球数目* @param curmaxv - 当前已经抓取小球中最大的编号* @param maxnum - 需要抓取小球的数目* @param maxv - 待抓取小球中最大的编号*/public static void kase3(int curnum, int curmaxv, int maxnum, int maxv){if(curnum == maxnum){cnt++;System.out.println(s);return;}for(int i = curmaxv + 1; i <= maxv; i++){ // i <= maxv - maxnum + curnum + 1s.push(i);kase3(curnum + 1, i, maxnum, maxv);s.pop();}}public static void main(String[] args){kase3(0, 0, 4, 8);System.out.println(cnt);}
}
输出：
[1, 2, 3, 4]
[1, 2, 3, 5]
[1, 2, 3, 6]
[1, 2, 3, 7]
[1, 2, 3, 8]
[1, 2, 4, 5]
[1, 2, 4, 6]
[1, 2, 4, 7]
[1, 2, 4, 8]
[1, 2, 5, 6]
[1, 2, 5, 7]
[1, 2, 5, 8]
[1, 2, 6, 7]
[1, 2, 6, 8]
[1, 2, 7, 8]
[1, 3, 4, 5]
[1, 3, 4, 6]
[1, 3, 4, 7]
[1, 3, 4, 8]
[1, 3, 5, 6]
[1, 3, 5, 7]
[1, 3, 5, 8]
[1, 3, 6, 7]
[1, 3, 6, 8]
[1, 3, 7, 8]
[1, 4, 5, 6]
[1, 4, 5, 7]
[1, 4, 5, 8]
[1, 4, 6, 7]
[1, 4, 6, 8]
[1, 4, 7, 8]
[1, 5, 6, 7]
[1, 5, 6, 8]
[1, 5, 7, 8]
[1, 6, 7, 8]
[2, 3, 4, 5]
[2, 3, 4, 6]
[2, 3, 4, 7]
[2, 3, 4, 8]
[2, 3, 5, 6]
[2, 3, 5, 7]
[2, 3, 5, 8]
[2, 3, 6, 7]
[2, 3, 6, 8]
[2, 3, 7, 8]
[2, 4, 5, 6]
[2, 4, 5, 7]
[2, 4, 5, 8]
[2, 4, 6, 7]
[2, 4, 6, 8]
[2, 4, 7, 8]
[2, 5, 6, 7]
[2, 5, 6, 8]
[2, 5, 7, 8]
[2, 6, 7, 8]
[3, 4, 5, 6]
[3, 4, 5, 7]
[3, 4, 5, 8]
[3, 4, 6, 7]
[3, 4, 6, 8]
[3, 4, 7, 8]
[3, 5, 6, 7]
[3, 5, 6, 8]
[3, 5, 7, 8]
[3, 6, 7, 8]
[4, 5, 6, 7]
[4, 5, 6, 8]
[4, 5, 7, 8]
[4, 6, 7, 8]
[5, 6, 7, 8]
70