问题描述:
输入一个复杂链表(每个节点中有节点值,以及两个指针,一个指向下一个节点,另一个特殊指针指向任意一个节点)。
方法1:直接复制法(消耗资源多)
首先遍历一遍原链表,创建新链表(赋值label和next),用map关联对应结点;再遍历一遍,更新新链表的random指针。
public RandomListNode Clone(RandomListNode pHead) {
if (pHead == null) {
return null;
}
RandomListNode p1 = pHead;
RandomListNode pHead2 = new RandomListNode(p1.label);
RandomListNode p2 = pHead2;
// 记录匹配节点
Map<RandomListNode, RandomListNode> map = new HashMap<RandomListNode, RandomListNode>();
map.put(pHead, pHead2);
while (p1 != null) {
if (p1.next != null) {
p2.next = new RandomListNode(p1.next.label);
} else {
p2.next = null;
}
map.put(p1.next, p2.next);
p1 = p1.next;
p2 = p2.next;
}
p1 = pHead;
p2 = pHead2;
while (p1 != null) {
p2.random = map.get(p1.random);
p1 = p1.next;
p2 = p2.next;
}
return pHead2;
}
方法2:
- 1、复制每个节点,如:复制节点A得到A1,将A1插入节点A后面
2、遍历链表,复制random节点 A1->random = A->random->next;
3、将链表拆分成原链表和复制后的链表
public RandomListNode Clone1(RandomListNode pHead){
if(pHead == null){
return null;
}
RandomListNode oldNode = pHead;
//复制每个节点,如:复制节点A得到A1,将A1插入节点A后面
while(oldNode != null){
RandomListNode newNode = new RandomListNode(oldNode.label);
newNode.next = oldNode.next;
oldNode.next = newNode;
oldNode = newNode.next;
}
//遍历链表,复制random节点 A1->random = A->random->next;
oldNode = pHead;
while(oldNode != null){
RandomListNode newNode = oldNode.next;
if(oldNode.random != null){
newNode.random = oldNode.random.next;
}
oldNode = newNode.next;
}
//将链表拆分成原链表和复制后的链表
RandomListNode newList = pHead.next;
RandomListNode temp;
oldNode = pHead;
while(oldNode.next != null){
temp = oldNode.next;
oldNode.next = temp.next;
oldNode = temp;
}
return newList;
}
class RandomListNode {
int label;
RandomListNode next = null;
RandomListNode random = null;
RandomListNode(int label) {
this.label = label;
}
}