2. Add Two Numbers
输入输出
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
答案一
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode c1 = l1;
ListNode c2 = l2;
ListNode sentinel = new ListNode(0);
ListNode d = sentinel;
int sum = 0;
while (c1 != null || c2 != null) {
sum /= 10;
if (c1 != null) {
sum += c1.val;
c1 = c1.next;
}
if (c2 != null) {
sum += c2.val;
c2 = c2.next;
}
d.next = new ListNode(sum % 10);
d = d.next;
}
if (sum / 10 == 1)
d.next = new ListNode(1);
return sentinel.next;
}
}
答案二
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
ListNode preHead(0), *p = &preHead;
int extra = 0;
while (l1 || l2 || extra) {
int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + extra;
extra = sum / 10;
p->next = new ListNode(sum % 10);
p = p->next;
l1 = l1 ? l1->next : l1;
l2 = l2 ? l2->next : l2;
}
return preHead.next;
}
总结
一个数，比如 15 ，15 / 10 = 1，15 % 10 = 5，前面得到的是十位上的数字，后面得到的是个位上的数字
参考
Is this Algorithm optimal or what? [c++] Sharing my 11-line c++ solution, can someone make it even more concise?
19. Remove Nth Node From End of List
输入输出
Given linked list: 1->2->3->4->5, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5.
答案一
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode start = new ListNode(0);
ListNode slow = start, fast = start;
slow.next = head;
//Move fast in front so that the gap between slow and fast becomes n
for(int i=1; i<=n+1; i++) {
fast = fast.next;
}
//Move fast to the end, maintaining the gap
while(fast != null) {
slow = slow.next;
fast = fast.next;
}
//Skip the desired node
slow.next = slow.next.next;
return start.next;
}
答案二
class Solution
{
public:
ListNode* removeNthFromEnd(ListNode* head, int n)
{
ListNode** t1 = &head, *t2 = head;
for(int i = 1; i < n; ++i)
{
t2 = t2->next;
}
while(t2->next != NULL)
{
t1 = &((*t1)->next);
t2 = t2->next;
}
*t1 = (*t1)->next;
return head;
}
};
总结
当涉及到链表尾的时候，要想到使用双指针来进行 next 操作
参考
Simple Java solution in one pass My short C++ solution
21. Merge Two Sorted Lists
答案一
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(l2->next, l1);
return l2;
}
}
};
答案二
class Solution {
public:
ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
ListNode dummy(INT_MIN);
ListNode *tail = &dummy;
while (l1 && l2) {
if (l1->val < l2->val) {
tail->next = l1;
l1 = l1->next;
} else {
tail->next = l2;
l2 = l2->next;
}
tail = tail->next;
}
tail->next = l1 ? l1 : l2;
return dummy.next;
}
};
总结
合并：递归：L->next = 调用函数()
参考
A recursive solution 14 line clean C++ Solution
23. Merge k Sorted Lists
答案一
public class Solution {
public ListNode mergeKLists(List<ListNode> lists) {
if (lists==null||lists.size()==0) return null;
PriorityQueue<ListNode> queue= new PriorityQueue<ListNode>(lists.size(),new Comparator<ListNode>(){
@Override
public int compare(ListNode o1,ListNode o2){
if (o1.val<o2.val)
return -1;
else if (o1.val==o2.val)
return 0;
else
return 1;
}
});
ListNode dummy = new ListNode(0);
ListNode tail=dummy;
for (ListNode node:lists)
if (node!=null)
queue.add(node);
while (!queue.isEmpty()){
tail.next=queue.poll();
tail=tail.next;
if (tail.next!=null)
queue.add(tail.next);
}
return dummy.next;
}
}
答案二
ListNode *mergeKLists(vector<ListNode *> &lists) {
if(lists.empty()){
return nullptr;
}
while(lists.size() > 1){
lists.push_back(mergeTwoLists(lists[0], lists[1]));
lists.erase(lists.begin());
lists.erase(lists.begin());
}
return lists.front();
}
ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
if(l1 == nullptr){
return l2;
}
if(l2 == nullptr){
return l1;
}
if(l1->val <= l2->val){
l1->next = mergeTwoLists(l1->next, l2);
return l1;
}
else{
l2->next = mergeTwoLists(l1, l2->next);
return l2;
}
}
总结
合并：堆
参考
A java solution based on Priority Queue Sharing my straightforward C++ solution without data structure other than vector
61. Rotate List
输入输出
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.
答案一
class Solution {
public:
ListNode* rotateRight(ListNode* head, int k) {
if(!head) return head;
int len=1; // number of nodes
ListNode *newH, *tail;
newH=tail=head;
while(tail->next) // get the number of nodes in the list
{
tail = tail->next;
len++;
}
tail->next = head; // circle the link
if(k %= len)
{
for(auto i=0; i<len-k; i++) tail = tail->next; // the tail node is the (len-k)-th node (1st node is head)
}
newH = tail->next;
tail->next = NULL;
return newH;
}
};
答案二
public ListNode rotateRight(ListNode head, int n) {
if (head==null||head.next==null) return head;
ListNode dummy=new ListNode(0);
dummy.next=head;
ListNode fast=dummy,slow=dummy;
int i;
for (i=0;fast.next!=null;i++)//Get the total length
fast=fast.next;
for (int j=i-n%i;j>0;j--) //Get the i-n%i th node
slow=slow.next;
fast.next=dummy.next; //Do the rotation
dummy.next=slow.next;
slow.next=null;
return dummy.next;
}
总结
参考
My clean C++ code, quite standard (find tail and reconnect the list) Share my java solution with explanation
82. Remove Duplicates from Sorted List II
移除所有重复的元素
输入输出
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.
答案一
public ListNode deleteDuplicates(ListNode head) {
if (head == null) return null;
if (head.next != null && head.val == head.next.val) {
while (head.next != null && head.val == head.next.val) {
head = head.next;
}
return deleteDuplicates(head.next);
} else {
head.next = deleteDuplicates(head.next);
}
return head;
}
答案二
public ListNode deleteDuplicates(ListNode head) {
if(head==null) return null;
ListNode FakeHead=new ListNode(0);
FakeHead.next=head;
ListNode pre=FakeHead;
ListNode cur=head;
while(cur!=null){
while(cur.next!=null&&cur.val==cur.next.val){
cur=cur.next;
}
if(pre.next==cur){
pre=pre.next;
}
else{
pre.next=cur.next;
}
cur=cur.next;
}
return FakeHead.next;
}
递归删除重复元素
参考
My accepted Java code My Recursive Java Solution
83. Remove Duplicates from Sorted List
移除重复的元素，但是还剩下一个哦
输入输出
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.
答案一
public ListNode deleteDuplicates(ListNode head) {
if(head == null || head.next == null)return head;
head.next = deleteDuplicates(head.next);
return head.val == head.next.val ? head.next : head;
}
答案二
public class Solution {
public ListNode deleteDuplicates(ListNode head) {
ListNode list = head;
while(list != null) {
if (list.next == null) {
break;
}
if (list.val == list.next.val) {
list.next = list.next.next;
} else {
list = list.next;
}
}
return head;
}
}