【算法22】-【合并两个排序的链表】
题目：输入两个递增排序的链表，合并这两个链表并使新链表中的节点仍然是递增排序的。
示例1：
输入：1->2->4, 1->3->4 输出：1->1->2->3->4->4
方法一：三指针实现，不引入头结点
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
// params check
if (l1 == null) {
return l2;
}
if (l2 == null) {
return l1;
}
// define pointer p1, p2 to traversal
ListNode p1 = l1;
ListNode p2 = l2;
// record head pointer
ListNode head = l1.val < l2.val ? l1 : l2;
// prev node
ListNode prev = head;
// compare the val of two node
while (p1 != null && p2 != null) {
if (p1.val < p2.val) {
if (prev != p1) {
prev.next = p1;
}
prev = p1;
ListNode temp = p1.next;
p1.next = p2;
p1 = temp;
} else {
if (prev != p2) {
prev.next = p2;
}
prev = p2;
ListNode temp = p2.next;
p2.next = p1;
p2 = temp;
}
}
return head;
}
} class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode dum = new ListNode(0), cur = dum;
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 != null ? l1 : l2;
return dum.next;
}
} 