代码随想录刷题-day04
代码随想录刷题day04
- 今日内容
-
- 两两交换链表中的节点
-
- 解题思路
- 实现
- 删除链表的倒数第N个结点
-
- 解题思路
- 实现
- 链表相交
-
- 解题思路
- 实现
- 环形链表II
-
- 解题思路
- 实现
今日内容
- 两两交换链表中的节点 19.删除链表的倒数第N个节点 面试题 02.07. 链表相交 142.环形链表II
两两交换链表中的节点
解题思路
- cur指向虚头节点（上一轮的最后节点），保存cur->next和下轮起始节点
- cur->next = 2;2->next = 1;1->next = next
实现
Java
class Solution {
public ListNode swapPairs(ListNode head) {
if(head == null || head.next == null){
return head;
}
ListNode next = head.next;
ListNode newNode = swapPairs(next.next);
next.next = head;
head.next = newNode;
return next;
}
}
C++
public:
ListNode* swapPairs(ListNode* head) {
ListNode* dummy = new ListNode(0,head);
ListNode* cur = dummy;
while(cur->next != nullptr && cur->next->next != nullptr){
ListNode* next = cur->next->next->next;
ListNode* tmp = cur->next;
cur->next = cur->next->next;
cur->next->next = tmp;
tmp->next = next;
cur = cur->next->next;
}
return dummy->next;
}
};
删除链表的倒数第N个结点
解题思路
- 使用一个虚头节点，使用双指针，快指针先走n，然后快慢一起走，慢指针指向被删节点的前节点
实现
Java
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode pre = new ListNode(0,head);
ListNode start = pre ,end = pre;
while(n!=0){
start = start.next;
n--;
}
while(start.next!=null){
start = start.next;
end = end.next;
}
end.next = end.next.next;
return pre.next;
}
}
C++
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* dummy = new ListNode(0,head);
ListNode* fast = dummy;
ListNode* slow = dummy;
while(n-- && fast != nullptr){
fast = fast->next;
}
while(fast->next != nullptr){
fast = fast->next;
slow = slow->next;
}
slow->next = slow->next->next;
return dummy->next;
}
};
链表相交
解题思路
- 两个链表一个遍历完了转向另一个，两个链表指向同一个元素，就是相交节点
实现
Java
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
ListNode A = headA;
ListNode B = headB;
while(A!=B){
A = A == null ? headB : A.next;
B = B == null ? headA : B.next;
}
return A;
}
}
C++
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
ListNode* A = headA;
ListNode* B = headB;
while(A!=B){
A = A==NULL ? headB : A->next;
B = B==NULL ? headA : B->next;
}
return A;
}
};
环形链表II
解题思路
- 双指针的方法，起点都是head，快指针走两步，慢指针走一步，快慢指针相遇表示是环形链表
- 快指针留在原位，慢指针回到head，两指针每次走一步，相遇节点表示是环的起始点。
实现
Java
public class Solution {
public ListNode detectCycle(ListNode head) {
ListNode fast = head;
ListNode slow = head;
while(fast!=null && fast.next!=null){
fast = fast.next.next;
slow = slow.next;
if(fast == slow){
ListNode index1 = fast;
ListNode index2 = head;
while(index1!=index2){
index1=index1.next;
index2=index2.next;
}
return index1;
}
}
return null;
}
}
C++
class Solution {
public:
ListNode *detectCycle(ListNode *head) {
ListNode* fast = head;
ListNode* slow = head;
while(fast != NULL && fast->next != NULL){
fast = fast->next->next;
slow = slow->next;
if(fast==slow){
ListNode* index1 = fast;
ListNode* index2 = head;
while(index1!=index2){
index1 = index1->next;
index2 = index2->next;
}
return index2;
}
}
return NULL;
}
};