Flatten the list so that all the nodes appear in a single-level, doubly linked list. You are given the head of the first level of the list.
题目： 430. Flatten a Multilevel Doubly Linked List
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分析：
当遇到有child的Node，先用*next存cur->next; 把cur->next指针指向child, child->pre指向cur； 接着以child为起点找到该曾最后一个Node，让它指向*next的节点。 接着以原来的child为cur Node重新开始一轮。
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| class Node {
public:
int val;
Node *prev;
Node *next;
Node *child;
Node() {}
Node(int _val, Node *_prev, Node *_next, Node *_child) {
val = _val;
prev = _prev;
next = _next;
child = _child;
}
};
Node *flatten(Node *head) {
Node *cur = head;
while (cur) {
if (cur->child) {
Node *next = cur->next;
cur->next = cur->child;
cur->child = nullptr;
cur->next->prev = cur;
Node *p = cur->next;
while (p->next) p = p->next;
p->next = next;
if (next) next->prev = p;
}
cur = cur->next;
}
return head;
}
|
时间复杂度： $O(n)$
