小问题

有没有大神感兴趣怎么实现，我自己实现的 code 不具备通用性，所以想参考下其他人的想法。目前问题纠结在，如何实现在遍历 list 的过程中还可能会删除多个元素的这个操作上

@frmongo 我用递归写了一个, 也仅仅在给出的这个上运行正常

# 实现了一个简单版的，不过有个假定：heads 和 tails 是各自唯一的。
# 如果不是唯一，可以对各个元素增加一个 count 字段，然后比较。

def merge(list_of_list):
heads = {a for (a, b) in list_of_list}
tails = {b for (a, b) in list_of_list}
return heads - tails

print(merge([[1,2],[2,3],[3,4],[4,7]])) # {1}
print(merge([[1,2],[2,3],[3,4],[4,7],[7,1]])) # set()

package temp;

import java.util.ArrayList;
import java.util.List;

public class D {

private List<Node> data = new ArrayList<>();

{
data.add(new Node(1, 2));
data.add(new Node(2, 3));
data.add(new Node(3, 4));
data.add(new Node(4, 7));
data.add(new Node(1, 7));
}

public static void main(String[] args) {
D start = new D();
List<Node> nn = start.check(start.data);
nn.forEach(n -> System.out.println(n.x + ":" + n.y));
}

private List<Node> check(List<Node> nodes){
if (nodes.size() == 1) {
return nodes;
}
List<Node> result = new ArrayList<>();
boolean can = false;
for (int i = 0; i < nodes.size(); i++) {
for (int j = i+1; j < nodes.size(); j++) {
if (nodes.get(i).y == nodes.get(j).x) {
result.add(new Node(nodes.get(i).x, nodes.get(j).y));
i++;
can = true;
break;
}else {
if (!result.contains(nodes.get(i))) {
result.add(nodes.get(i));
}
if (!result.contains(nodes.get(j))) {
result.add(nodes.get(j));
}
}
}
if (i == nodes.size() -1) {
if (!result.contains(nodes.get(i))) {
result.add(nodes.get(i));
}
}
}
if (can)
result = check(result);
return result;
}

class Node{
Node(int x, int y) {
this.x = x;
this.y = y;
}

int x;
int y;
}

}

我写了一个，但是 fusion 是一个有缺陷的函数，需要改进

from random import choice

def combine_two(a,b):
c = [None,None]
if a[1]==b[0]:
c[0]=a[0]
c[1]=b[1]
elif b[1] == a[0]:
c[0]=b[0]
c[1]=a[1]
return c

def fusion(list):
ss = list[:]
nn = len(ss)
for i in range(100):
ele_1 = choice(ss)
ele_2 = choice(ss)
c = combine_two(ele_1,ele_2)
if c != [None,None]:
ss.remove(ele_1)
ss.remove(ele_2)
ss.append(c)
return ss

def check(list):
n = len(list)
for i in range(n):
for j in range(n):
if i != j:
if list[i][0] == list[j][0] and list[i][1] == list[j][1]:
return False
return True

jj = [[2,3],[3,4],[1,2],[4,7],[11,13],[1,7]]
m = fusion(jj)
print m
n = check(m)
print n

```python
list_a = [[1,2],[2,3],[3,4],[4,7],[1,7]]
for i in list_a:
list_test = [(i[0],x[1]) for x in list_a if i != x and i[1] == x[0]]
if list_test:
print(i, list_test)
```