标签：head ListNode val next 链表 l2 NULL LeetCode
2. Add Two Numbers：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* ret, *tmp;
int carry = 0, sum=0;
ret = tmp = new ListNode(0);
while(carry||l1||l2){
sum = carry;
if(l1)sum+=l1->val;
if(l2)sum+=l2->val;
carry = sum/10;
tmp->val = sum%10;
if(l1)l1 = l1->next?l1->next:NULL;
if(l2)l2 = l2->next?l2->next:NULL;
if(!l1 && !l2 && !carry)return ret;
tmp->next = new ListNode(0);
tmp=tmp->next;
}
return NULL;
}
};
19. Remove Nth Node From End of List
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
if(head == NULL || n == 0) return head;
ListNode* slow = head;
ListNode* fast = head;
ListNode* pre = head;
if(fast->next == NULL) return NULL; //[1] 1
while(n>0){ //if n, slow will just point to the nth
fast = fast->next;
n--;
}
while(fast != NULL){
fast = fast->next;
pre = slow;
slow = slow->next;
}
pre->next = slow->next;
//if pre==slow==head [1,2] 2
if(pre == slow) return head->next;
return head;
}
};
21. Merge Two Sorted Lists
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode *retList = NULL, *tempList = NULL;
if(!l1 && !l2) return retList;
int val = 0, val1 = 0, val2 = 0;
retList = tempList = new ListNode(0);
while(l1 || l2){
val1 = l1?l1->val:0;
val2 = l2?l2->val:0;
if(((val1<val2) && l1) || !l2){
tempList->val = val1;
if(!(l1->next) && !l2) return retList;
tempList->next = new ListNode(0);
tempList = tempList->next;
l1 = (l1->next)?l1->next:NULL;
continue;
}
else{
tempList->val = val2;
if(!(l2->next) && !l1) return retList;
tempList->next = new ListNode(0);
tempList = tempList->next;
l2 = (l2->next)?l2->next:NULL;
continue;
}
}
return retList;
}
//Other guy's solution
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2)
{
if(!l1) // If no l1, return l2
return l2;
if(!l2) // If no l2, return l1
return l1;
if(!l2 && !l1) // If neither, return NULL;
return NULL;
ListNode* head; // The pointer we will use to construct a merged list
if(l1->val < l2->val) // If l1 less than l2
{
head = l1; // We start at l1
l1 = l1->next; // and iterate l1
}
else // If l2 less than l1
{
head = l2; // We start at l2
l2 = l2->next; // and iterate l2
}
ListNode* ret = head; // We need to save the addres of the head of the list
while(l1 && l2) // While both input lists have values
{
if(l1->val < l2->val) // Compare the current values, if l1 is less
{
head->next = l1; // Append the merged list with l1's current address
l1 = l1->next; // Advance l1
}
else // Else, l2 had the low value
{
head->next = l2; // Append l2 to the list
l2 = l2->next; // Advance l2
}
head->next->next = NULL; // Append a NULL teminator to the list
head = head->next; // Advance the merged list
}
// Lastly, if list were different lengths, we need to append the longer list tail to the merged list
if(l1)
head->next = l1;
else if(l2)
head->next = l2;
return ret; // Return the starting address of head that we saved.
}
24. Swap Nodes in Pairs
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* swapPairs(ListNode* head) {
ListNode* first = head;
if(head == NULL || head->next == NULL)return head;
ListNode* second = head->next;
ListNode* ret = second;
ListNode* third = second->next;
second->next = first;
while(third && third->next){
first->next = third->next;
first = third;
second = third->next;
third = second->next;
second->next = first;
}
if(third == NULL){
second->next = first;
first->next = NULL;
}else{
//third->next =NULL
first->next = third;
}
return ret;
}
};
61. Rotate List
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* rotateRight(ListNode* head, int k) {
if(head == NULL || k==0) return head;
int len = 1;
ListNode* orighead = head;
ListNode* newhead=head;
while(head->next){
len++;
head = head->next;
}
//loop the link
head->next = orighead;
k = len - k%len-1; //find the front node of kth.
while(k){
newhead = newhead->next;
k--;
}
ListNode* ret = newhead->next;
newhead->next = NULL;
return ret;
}
};
83. Remove Duplicates from Sorted List：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head) {
ListNode *dup = NULL;
ListNode *cur = head;
if(cur == NULL || cur->next == NULL) return head;
ListNode *nxt = cur->next;
while(nxt != NULL){
if(cur->val == nxt->val){
ListNode *dup = nxt;
nxt = nxt->next;
cur->next = nxt;
delete dup;
}else{
cur = nxt;
nxt = nxt->next;
}
}
return head;
}
};
141. Linked List Cycle
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
bool hasCycle(ListNode *head) {
ListNode* slow = head;
ListNode* fast = head;
while(fast && fast->next){
slow = slow->next;
fast = fast->next->next; //2step
if(slow == fast) return true;
}
return false;
}
};
147. Insertion Sort List
class Solution {
public:
ListNode* insertionSortList(ListNode* head) {
ListNode* dummy = new ListNode(0);
ListNode* pre = dummy;
ListNode* curr = head;
if(head == NULL || head->next == NULL) return head;
ListNode* next = NULL;
while(curr != NULL){
next = curr->next;
while(pre->next != NULL && pre->next->val < curr->val){ //find insert pos
pre = pre->next;
}
curr->next = pre->next; //insert
pre->next = curr;
curr = next; //move curr to next node
pre = dummy;//reset the pre
}
return dummy->next;
}
};
148. Sort List
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* sortList(ListNode* head) {
ListNode*p = head;
ListNode*q = NULL;
int temp = 0;
if(head == NULL) return head;
for(; p!=NULL; p=p->next)
for(q=p->next; q!=NULL; q=q->next){
if(p->val > q->val){
temp = p->val;
p->val = q->val;
q->val = temp;
}
}
return head;
}
};
160. Intersection of Two Linked Lists
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
ListNode* p1 = headA;
ListNode* p2 = headB;
while(p1 != p2){
p1 = p1?p1->next:headB;
p2 = p2?p2->next:headA;
}
return p1;
}
};
203. Remove Linked List Elements
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeElements(ListNode* head, int val) {
if(head == NULL) return head;
ListNode* p = head;
while (p->next != NULL){
if(p->next->val == val){
ListNode *tmp = p->next;
p->next = p->next->next;
delete tmp;
}
else{
p = p->next;
}
}
if(head->val == val)
head = head->next;
return head;
}
};
206. Reverse Linked List：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if(head == NULL || head->next == NULL) return head;
ListNode* pre = NULL;
ListNode* cur = head;
ListNode* nx = cur->next;
while(nx != NULL){
cur->next = pre;
pre = cur;
cur = nx;
nx = nx->next;
}
cur->next = pre;
return cur;
}
};
237. Delete Node in a Linked List
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
void deleteNode(ListNode* node) {
ListNode* p = node->next;
node->val = p->val;
node->next = p->next;
delete p;
}
};
https://www.cnblogs.com/upcwanghaibo/p/6928887.html
https://blog.csdn.net/qq_37466121/article/details/88204678
https://www.cnblogs.com/upcwanghaibo/p/6928887.html
标签：head,ListNode,val,next,链表,l2,NULL,LeetCode 来源： https://www.cnblogs.com/feliz/p/11147347.html
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