给定一个大小为 n 的数组,生成并打印数组中 r 元素的所有可能排列。
例如,如果 n 为 4,输入数组为 [0, 1, 2, 3],r 为 3,则输出应为
[0, 1, 2]
[0, 1, 3]
[0, 2, 1]
[0, 2, 3]
[0, 3, 1]
[0, 3, 2]
[1, 0, 2]
[1, 0, 3]
[1, 2, 0]
[1, 2, 3]
[1, 3, 0]
[1, 3, 2]
[2, 0, 1]
[2, 0, 3]
[2, 1, 0]
[2, 1, 3]
[2, 3, 0]
[2, 3, 1]
[3, 0, 1]
[3, 0, 2]
[3, 1, 0]
[3, 1, 2]
[3, 2, 0]
[3, 2, 1]
输入数组中的所有元素都是整数,从 0 到 n-1。如何使用 Java 打印所有可能的排列?
重要提示:单个排列中所有元素的数量并不总是原始数组的大小。它小于或等于原始数组的大小。此外,每个排列中元素的顺序也很重要。
我在n=4和r=3时编写了一些代码。如果 n = 100 且 r = 50,我的代码将非常难看。当参数只有 n 和 r 时,有什么聪明的方法可以做到这一点吗?
public static void main(String[] args) {
// original array
ArrayList < Integer > items = new ArrayList < > ();
// all permutations
ArrayList < ArrayList < Integer >> allPermutations = new ArrayList < ArrayList < Integer >> ();
// define the end item of the original array.
// this is n, suppose it's 4 now.
int endItem = 4;
for (int i = 0; i < endItem; i++) {
items.add(i);
}
// r means how many elements in each single permutation
// suppose it's 3 now, i have to write for-loop three times
// if r is 100, i have to write for-loop 100 times
// first of the "r"
for (int i = 0; i < items.size(); i++) {
// second of the "r"
for (int j = 0; j < items.size(); j++) {
// can't be identical to i
if (j == i)
continue;
// third of the "r"
for (int k = 0; k < items.size(); k++) {
// can't be identical to i or j
if (k == i || k == j)
continue;
// a single permutation
ArrayList < Integer > singlePermutation = new ArrayList < > ();
singlePermutation.add(items.get(i));
singlePermutation.add(items.get(j));
singlePermutation.add(items.get(k));
// all permutations
allPermutations.add(singlePermutation);
}
}
}
for (ArrayList < Integer > permutation: allPermutations) {
System.out.println(permutation);
}
System.out.println(allPermutations.size());
}
最佳答案
将解决方案从问题移至答案:
Solution:
Thanks to older coder, I managed to find the solution.
public class PermutationTest10 { // a is the original array // k is the number of elements in each permutation public static ArrayList<ArrayList<Integer>> choose(ArrayList<Integer> a, int k) { ArrayList<ArrayList<Integer>> allPermutations = new ArrayList<ArrayList<Integer>>(); enumerate(a, a.size(), k, allPermutations); return allPermutations; } // a is the original array // n is the array size // k is the number of elements in each permutation // allPermutations is all different permutations private static void enumerate(ArrayList<Integer> a, int n, int k, ArrayList<ArrayList<Integer>> allPermutations) { if (k == 0) { ArrayList<Integer> singlePermutation = new ArrayList<Integer>(); for (int i = n; i < a.size(); i++){ singlePermutation.add(a.get(i)); } allPermutations.add(singlePermutation); return; } for (int i = 0; i < n; i++) { swap(a, i, n-1); enumerate(a, n-1, k-1, allPermutations); swap(a, i, n-1); } } // helper function that swaps a.get(i) and a.get(j) public static void swap(ArrayList<Integer> a, int i, int j) { Integer temp = a.get(i); a.set(i, a.get(j)); a.set(j, temp); } // sample client public static void main(String[] args) { // n is the end item of the array. // if n = 5, the array is [0, 1, 2, 3, 4, 5] // k is the number of elements of each permutation. int n =5; int k =3; // create original array ArrayList<Integer> elements = new ArrayList<> (); for (int i =0; i < n; i ++){ elements.add(i); } ArrayList<Integer> a = new ArrayList<> (); for (int i = 0; i < n; i ++){ a.add(elements.get(i)); } System.out.println(choose(a, k)); } }
关于java - 在 Java 中打印给定大小为 n 的整数数组中 r 元素的所有可能排列,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44949030/