剑指offer(12)

Posted 2021-02-11 figsprite

来两道关于链表链接的题目：

题目一：

　　输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。

　　本题要考虑到其中一条链表是空或者两个都是空的情况。

　　在每个链表安上一个指针，对比一次，提取一个结点，接到目标链表上。

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode Merge(ListNode list1,ListNode list2) {
        if(list1==null)
            return list2;
        if(list2==null)
            return list1;
        
        ListNode listMergeHead = null;
        ListNode point1 = list1;
        ListNode point2 = list2;
        
        if(point1.val<point2.val){
            listMergeHead = point1;
            listMergeHead.next = Merge(point1.next,point2);
        }else{
            listMergeHead = point2;
            listMergeHead.next = Merge(point1,point2.next);
        }
        
        return listMergeHead;
    }
}

　　接下来给出非递归方式

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode Merge(ListNode list1,ListNode list2) {
        if(list1==null)
            return list2;
        if(list2==null)
            return list1;
        
        ListNode listMergeHead = null;
        ListNode current = null;
        
        while(list1!=null&&list2!=null){
            if(list1.val<list2.val){
                if(listMergeHead==null){
                    listMergeHead=current=list1;
                }else{
                    current.next=list1;
                    current = current.next;
                }
                list1=list1.next;
            }else{
                if(listMergeHead==null){
                    listMergeHead=current=list2;
                }else{
                    current.next=list2;
                    current = current.next;
                }
                list2=list2.next;
            }
        }
        
        if(list1==null){
            current.next=list2;
        }
        if(list2==null){
            current.next=list1;
        }
        return listMergeHead;
    }
}