标签:pre 有序 ext def ted 递归 app 拼接 有序链表合并
问题描述: 将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
示例:
输入:1->2->4, 1->3->4
输出:1->1->2->3->4->4
迭代:使用双指针分别指向l1和l2,比较出较小值结点
//C
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2){
if(!l1) return l2;
if(!l2) return l1;
struct ListNode *p1, *p2, *pa, *head;
p1 = l1;
p2 = l2;
if(p1-> val < p2-> val){
head = p1;
p1 = p1 -> next;
}
else {
head = p2;
p2 = p2 -> next;
}
pa = head;
while(p1 != NULL && p2 != NULL){
if(p1 -> val <= p2 -> val){
pa -> next = p1;
pa = p1;
p1 = p1 -> next;
}
else{
pa -> next = p2;
pa = p2;
p2 = p2 -> next;
}
}
if(p1 != NULL) pa -> next = p1;
if(p2 != NULL) pa -> next = p2;
return head;
}
//JS
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var mergeTwoLists = function(l1, l2) {
if(!l1) return l2;
if(!l2) return l1;
let p = l1, q = l2, head, pa;
if(p.val < q.val) {
head = p;
p = p.next;
}
else {
head = q;
q = q.next;
}
pa = head;
while(p && q){
if(p.val < q.val) {
pa.next = p;
pa = p;
p = p.next;
}
else{
pa.next = q;
pa = q;
q = q.next;
}
}
if(p) pa.next = p;
if(q) pa.next = q;
return head;
};
递归:
//C
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2){
if(l1==NULL){
return l2;
}
if(l2==NULL){
return l1;
}
if(l1->val<l2->val || l1->val==l2->val){
l1->next=mergeTwoLists(l1->next,l2);return l1;
}
else{
l2->next=mergeTwoLists(l1,l2->next);return l2;
}
}
//JS
var mergeTwoLists = function(l1, l2) {
if(!l1) return l2;
if(!l2) return l1;
if(l1.val < l2.val){
l1.next = mergeTwoLists(l1.next, l2);
return l1;
}
else{
l2.next = mergeTwoLists(l1, l2.next);
return l2;
}
};
标签:pre 有序 ext def ted 递归 app 拼接 有序链表合并
原文地址:https://www.cnblogs.com/JesseyWang/p/12968951.html