复杂链表的复制(java)

题目

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),返回结果为复制后复杂链表的head。(注意,输出结果中请不要返回参数中的节点引用,否则判题程序会直接返回空)

分析
(1),1 -> 2 -> 3
(2),1 -> 1(copy) -> 2 ->2(copy) -> 3 -> 3 (copy)
(3),设置每个复制节点的random节点
(4),分离出新的链表即可

代码实现

/*
public class RandomListNode {
    int label;
    RandomListNode next = null;
    RandomListNode random = null;

    RandomListNode(int label) {
        this.label = label;
    }
}
*/
public class Solution {
    public RandomListNode Clone(RandomListNode pHead)
    { 
        if(pHead == null){
            return pHead;
        }
        RandomListNode cur = pHead;
        RandomListNode next = null;
        while(cur != null){
            RandomListNode tmp = new RandomListNode(cur.label);
            next = cur.next;
            cur.next = tmp;
            tmp.next = next;
            cur = next;
        }
        cur = pHead;
        while(cur != null){
            next = cur.next.next;
            if(cur.random != null){
                cur.next.random = cur.random.next;
            }
            cur = next;
        }
        RandomListNode res = pHead.next;
        RandomListNode curRes = res;
        cur = pHead;
        while(cur != null){
            cur.next = cur.next.next;
            if(curRes.next != null){
                curRes.next = curRes.next.next;
            }
            cur = cur.next;
            curRes = curRes.next;
        }
        return res;
    }
}

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