题目描述
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。
解题思路
- 生成一个新的头节点
- 最后返回头节点的next指针
常规版本
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
if(pHead1 == NULL && pHead2 == NULL){
return NULL;
}
if(pHead1 == NULL){
return pHead2;
}
if(pHead2 == NULL){
return pHead1;
}
ListNode* p = pHead1, *q = pHead2;
ListNode* merge_head = new ListNode(-1); //生成一个新的头节点，头结点的next指向合并后的第一个节点，最后返回时，返回头结点的next即可
if(p -> val <= q -> val){
merge_head -> next = p;
p = p -> next;
}else{
merge_head -> next = q;
q = q -> next;
}
ListNode* tail = merge_head -> next;
while(p && q){
if(p -> val <= q -> val){
tail -> next = p;
tail = p;
p = p -> next;
}else{
tail -> next = q;
tail = q;
q = q -> next;
}
}
if(p != NULL){
tail -> next = p;
}
if(q != NULL){
tail -> next = q;
}
return merge_head -> next;
}
};
递归版本
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
if(pHead1 == NULL && pHead2 == NULL){
return NULL;
}
if(pHead1 == NULL){
return pHead2;
}
if(pHead2 == NULL){
return pHead1;
}
ListNode* merge_head = NULL;
if(pHead1 -> val <= pHead2 -> val){
merge_head = pHead1;
merge_head -> next = Merge(pHead1 -> next, pHead2);
}else{
merge_head = pHead2;
merge_head ->next = Merge(pHead1, pHead2 -> next);
}
return merge_head;
}
};
第二遍问题
- ListNode* p = newNode -> next;
- p -> next = pNode1
- 然后就要移动p指针
- 最后才移动pNode指针
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode* newNode = new ListNode(-1);
if(l1 == NULL && l2 == NULL){
return NULL;
}
if(l1 == NULL && l2 != NULL){
return l2;
}
if(l1 != NULL && l2 == NULL){
return l1;
}
ListNode* pNode1 = l1;
ListNode* pNode2 = l2;
if(pNode1 -> val < pNode2 -> val){
newNode -> next = pNode1;
pNode1 = pNode1 -> next;
}else{
newNode -> next = pNode2;
pNode2 = pNode2 -> next;
}
ListNode* p = newNode -> next;
while(pNode1 != NULL && pNode2 != NULL){
if(pNode1 -> val <= pNode2 -> val){
p -> next = pNode1;
p = pNode1;
pNode1 = pNode1 -> next;
}else{
p -> next = pNode2;
p = pNode2;
pNode2 = pNode2 -> next;
}
}
if(pNode1){
p -> next = pNode1;
}
if(pNode2){
p -> next = pNode2;
}
return newNode -> next;
}
};
