以所有可能的方式將列表拆分為兩個子列表

Question

我有一個可變大小的列表，例如

[1, 2, 3, 4]

我想盡一切可能把這個列表分成兩部分：

([], [1, 2, 3, 4])
([1], [2, 3, 4])
([2], [1, 3, 4])
([3], [1, 2, 4])
([4], [1, 2, 3])
([1, 2], [3, 4])
([1, 3], [2, 4])
([1, 4], [2, 3])
([2, 3], [1, 4])
([2, 4], [1, 3])
([3, 4], [1, 2])
([1, 2, 3], [4])
([1, 2, 4], [3])
([1, 3, 4], [2])
([2, 3, 4], [1])
([1, 2, 3, 4], [])

我很確定這不是一個未知問題，並且可能有一種算法可以解決這個問題，但是我找不到。 此外，這不應該使用任何外部庫，而是使用大多數語言中的簡單語言特性（循環、條件、方法/函數、變量......）。

我在 Python 中編寫了一個 hackish 解決方案：

def get_all(objects):
    for i in range(1, len(objects)):
        for a in combinations(objects, i):
            for b in combinations([obj for obj in objects if obj not in up], len(objects) - i):
                yield State(up, down)
    if objects:
        yield State([], objects)
        yield State(objects, [])

但是，它使用庫功能，一般看起來不是很好。

Answer 1

l = [1, 2, 3, 4]
flags = [False] * len(l)
while True:
    a = [l[i] for i, flag in enumerate(flags) if flag]
    b = [l[i] for i, flag in enumerate(flags) if not flag]
    print a, b
    for i in xrange(len(l)):
        flags[i] = not flags[i]
        if flags[i]:
            break
    else:
        break

結果：

[] [1, 2, 3, 4]
[1] [2, 3, 4]
[2] [1, 3, 4]
[1, 2] [3, 4]
[3] [1, 2, 4]
[1, 3] [2, 4]
[2, 3] [1, 4]
[1, 2, 3] [4]
[4] [1, 2, 3]
[1, 4] [2, 3]
[2, 4] [1, 3]
[1, 2, 4] [3]
[3, 4] [1, 2]
[1, 3, 4] [2]
[2, 3, 4] [1]
[1, 2, 3, 4] []

它可以很容易地適應java：

public static void main(String[] args) {
    int[] l = new int[] { 1, 2, 3, 4 };
    boolean[] flags = new boolean[l.length];
    for (int i = 0; i != l.length;) {
        ArrayList<Integer> a = new ArrayList<>(), b = new ArrayList<>();
        for (int j = 0; j < l.length; j++)
            if (flags[j]) a.add(l[j]); else b.add(l[j]);
        System.out.println("" + a + ", " + b);
        for (i = 0; i < l.length && !(flags[i] = !flags[i]); i++);
    }
}

Answer 2

使用按位算術計算應易於轉換為Java的子集的更低級別的解決方案：

def sublists(xs):
    l = len(xs)
    for i in range(1 << l):
        incl, excl = [], []
        for j in range(l):
            if i & (1 << j):
                incl.append(xs[j])
            else:
                excl.append(xs[j])
        yield (incl, excl)

Answer 3

雖然在Python中，使用其豐富的庫來獲取結果非常容易，但在Java中，您可以編寫遞歸解決方案。 以下將打印陣列的所有可能組合：

public static void main(String[] args) {
    List<Integer> num = Arrays.asList(1, 2, 3, 4);
    List<List<Integer>> sublists = new ArrayList<List<Integer>>();
    for (int i = 0; i <= num.size(); i++) {
      permutation(num, sublists, i, new ArrayList<Integer>(), 0);
    }

    for (List<Integer> subList : sublists) {
        List<Integer> numCopy = new ArrayList<Integer>(num);
        numCopy.removeAll(subList);
        System.out.println("(" + subList + ", " + numCopy + ")");
    }
}

public static void permutation(List<Integer> nums, List<List<Integer>> subLists, int sublistSize, List<Integer> currentSubList,
      int startIndex) {
    if (sublistSize == 0) {
      subLists.add(currentSubList);
    } else {
      sublistSize--;
      for (int i = startIndex; i < nums.size(); i++) {
        List<Integer> newSubList = new ArrayList<Integer>(currentSubList);
        newSubList.add(nums.get(i));
        permutation(nums, subLists, sublistSize, newSubList, i + 1);
      }
    }
}

sublists包含到目前為止發現的所有組合。 最后一個參數是當前子列表的下一個元素的startIndex 。 那是為了避免重復。

Answer 4

遍歷所有不同大小的組合並從原始列表中“減去”它們似乎是直觀的方法IMO：

from itertools import combinations

s = [1, 2, 3, 4]
for combs in (combinations(s, r) for r in range(len(s)+1))  :
    for comb in combs:
        diff = list(set(s[:]) - set(comb))
        print diff, list(comb)

OUTPUT

[1, 2, 3, 4] []
[2, 3, 4] [1]
[1, 3, 4] [2]
[1, 2, 4] [3]
[1, 2, 3] [4]
[3, 4] [1, 2]
[2, 4] [1, 3]
[2, 3] [1, 4]
[1, 4] [2, 3]
[1, 3] [2, 4]
[1, 2] [3, 4]
[4] [1, 2, 3]
[3] [1, 2, 4]
[2] [1, 3, 4]
[1] [2, 3, 4]
[] [1, 2, 3, 4]

Java可以應用相同的方法（只是它更冗長......）：

private static List<Integer> initial;

public static void main(String[] args) throws IOException {
    initial = Arrays.asList(1, 2, 3);
    combinations(initial);
}

static void combinations(List<Integer> src) {
    combinations(new LinkedList<>(), src);
}

private static void combinations(LinkedList<Integer> prefix, List<Integer> src) {        
    if (src.size() > 0) {
        prefix = new LinkedList<>(prefix); //create a copy to not modify the orig
        src = new LinkedList<>(src); //copy
        Integer curr = src.remove(0);
        print(prefix, curr); // <-- this is the only thing that shouldn't appear in a "normal" combinations method, and which makes it print the list-pairs
        combinations(prefix, src); // recurse without curr
        prefix.add(curr);
        combinations(prefix, src); // recurse with curr
    }
}

// print the prefix+curr, as one list, and initial-(prefix+curr) as a second list
private static void print(LinkedList<Integer> prefix, Integer curr) {
    prefix = new LinkedList<>(prefix); //copy
    prefix.add(curr);
    System.out.println(Arrays.toString(prefix.toArray()) +
                    " " + Arrays.toString(subtract(initial, prefix).toArray()));
}

private static List<Integer> subtract(List<Integer> initial, LinkedList<Integer> prefix) {
    initial = new LinkedList<>(initial); //copy
    initial.removeAll(prefix);
    return initial;
}

OUTPUT

[1] [2, 3]
[2] [1, 3]
[3] [1, 2]
[2, 3] [1]
[1, 2] [3]
[1, 3] [2]
[1, 2, 3] []

Answer 5

我發現其他解決方案很難理解，這些問題可以通過遞歸輕松解決：

alist = [1, 2, 3, 4]

def partition(alist, left, right, index):
    if index >= len(alist):
        print(left, right)
        return
    partition(alist, left + [alist[index]], right, index + 1)
    partition(alist, left, right + [alist[index]], index + 1)


print(partition(alist, [], [], 0))

或使用生成器：

def partition(alist, left, right, index):
    if index >= len(alist):
        yield left, right
    else:
        yield from partition(alist, left + [alist[index]], right, index + 1)
        yield from partition(alist, left, right + [alist[index]], index + 1)

print(list(partition(alist, [], [], 0)))

Output：

[1, 2, 3, 4] []
[1, 2, 3] [4]
[1, 2, 4] [3]
[1, 2] [3, 4]
[1, 3, 4] [2]
[1, 3] [2, 4]
[1, 4] [2, 3]
[1] [2, 3, 4]
[2, 3, 4] [1]
[2, 3] [1, 4]
[2, 4] [1, 3]
[2] [1, 3, 4]
[3, 4] [1, 2]
[3] [1, 2, 4]
[4] [1, 2, 3]
[] [1, 2, 3, 4]