LinkedList - 21. Merge Two Sorted Lists

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  1. Merge Two Sorted Lists

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

思路:

合并两个有序链表,做法有两种,递归和迭代,递归的条件就是返回两个节点中小的那个节点。迭代就是正常遍历,然后比较两个链表的节点,生成新的合并链表。

代码:

golang:

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
    var res, tmp *ListNode
    
    for l1 != nil && l2 != nil {
        if l1.Val < l2.Val {
            res, tmp, l1 = connect(res, tmp, l1)
        } else {
            res, tmp, l2 = connect(res, tmp, l2)
        }
    }
    
    res, tmp = fill(res, tmp, l1)
    res, tmp = fill(res, tmp, l2)
    return res
}

func connect(res, tmp, l *ListNode) (*ListNode, *ListNode, *ListNode) {
    if res == nil {
        res = l
        tmp = l
    } else {
        tmp.Next = l // tmp后接l这个节点
        tmp = l // tmp 后移一位
    }
    l = l.Next
    return res, tmp, l
}

func fill(res, tmp, l *ListNode) (*ListNode, *ListNode) {
    if l != nil {
        if res == nil {
            res = l
        } else {
            tmp.Next = l
        }
    }
    return res, tmp
}

java:

/**

 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution { 
   /* // recursion
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        
        if (l1.val < l2.val) l1.next = mergeTwoLists(l1.next, l2);
        else l2.next = mergeTwoLists(l1, l2.next);
        
        return l1.val < l2.val ? l1 : l2;
    }*/
    
    // iteratively
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode result = null;
        ListNode temp = null; 
        
        while(l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                if (result == null) {
                    result = l1;
                    temp = l1;
                } else {
                    temp.next = l1;
                    temp = l1;
                }
                l1 = l1.next;
            } else {
                if (result == null) {
                    result = l2;
                    temp = l2;
                } else {
                    temp.next = l2;
                    temp = l2;
                }
                l2 = l2.next;
            }
        }
        
        if (l1 != null) {
            if (result == null) {
                result = l1;
            } else {
                temp.next = l1;
            }
        
        } else {
            if (result == null) {
                result = l2;
            } else {
                temp.next = l2;
            }
        }
        
        return result;
    }
}