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Merge Two Sorted Lists (E)

题目

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

题意

将两个有序链表合并为一个有序链表。

思路

归并排序。

代码实现

Java

迭代

class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode head = new ListNode(0);
        ListNode pointer = head;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                pointer.next = l1;
                l1 = l1.next;
            } else {
                pointer.next = l2;
                l2 = l2.next;
            }
            pointer = pointer.next;
        }
        if (l1 == null) {
            pointer.next = l2;
        }
        if (l2 == null) {
            pointer.next = l1;
        }
        return head.next;
    }
}

递归

class Solution {
    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);
            return l1;
        } else {
            l2.next = mergeTwoLists(l1, l2.next);
            return l2;
        }
    }
}

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var mergeTwoLists = function (l1, l2) {
  let dummy = new ListNode(0)
  let cur = dummy
  while (l1 !== null && l2 !== null) {
    if (l1.val < l2.val) {
      cur.next = l1
      l1 = l1.next
    } else {
      cur.next = l2
      l2 = l2.next
    }
    cur = cur.next
  }
  cur.next = l1 ? l1 : l2
  return dummy.next
}

0021. Merge Two Sorted Lists (E)

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原文地址：https://www.cnblogs.com/mapoos/p/13171280.html