给定一组 {1,2,3,4,5 ... N}
n个元素,我们需要找到一个长度为k的所有子集。
Given a set {1,2,3,4,5...n}
of n elements, we need to find all subsets of length k .
例如,如果n = 4和k = 2,输出
是 {1,2},{1,3}, {1,4},{2,3},{2,4},{3,4}
For example, if n = 4 and k = 2, the output
would be {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}
.
我甚至无法弄清楚如何下手。我们没有使用内置的库函数像next_permutation等
I am not even able to figure out how to start. We don't have to use the inbuilt library functions like next_permutation etc.
需要的算法和实现用C / C ++或Java。
Need the algorithm and implementation in either C/C++ or Java.
推荐答案
递归是你的朋友对这个任务。
Recursion is your friend for this task.
有关每个元素 - 猜测,如果它是在当前子集,并且与猜测递归调用和一个较小的超集可以从选择。这样做的同时为是和否猜测 - 将导致所有可能子集。 制约自己到一定长度可在停机条款很容易做到。
For each element - "guess" if it is in the current subset, and recursively invoke with the guess and a smaller superset you can select from. Doing so for both the "yes" and "no" guesses - will result in all possible subsets. Restraining yourself to a certain length can be easily done in a stop clause.
Java的code:
Java code:
private static void getSubsets(List<Integer> superSet, int k, int idx, Set<Integer> current,List<Set<Integer>> solution) {
//successful stop clause
if (current.size() == k) {
solution.add(new HashSet<>(current));
return;
}
//unseccessful stop clause
if (idx == superSet.size()) return;
Integer x = superSet.get(idx);
current.add(x);
//"guess" x is in the subset
getSubsets(superSet, k, idx+1, current, solution);
current.remove(x);
//"guess" x is not in the subset
getSubsets(superSet, k, idx+1, current, solution);
}
public static List<Set<Integer>> getSubsets(List<Integer> superSet, int k) {
List<Set<Integer>> res = new ArrayList<>();
getSubsets(superSet, k, 0, new HashSet<Integer>(), res);
return res;
}
与调用:
List<Integer> superSet = new ArrayList<>();
superSet.add(1);
superSet.add(2);
superSet.add(3);
superSet.add(4);
System.out.println(getSubsets(superSet,2));
将产生:
[[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]]
相关推荐
最新文章