代码随想录刷题第四天 | 24. 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

24. 两两交换链表中的节点

有一个点学到了 while循环那里的cur->next!=nullptr&&cur->next->next!=nullptr 顺序不要错了

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* swapPairs(ListNode* head) {ListNode *dummyhead  = new ListNode(0);dummyhead->next = head;ListNode *cur = dummyhead;while(cur->next!=nullptr&&cur->next->next!=nullptr){ListNode *temp = cur->next;ListNode *temp1 = cur->next->next->next;cur->next = cur->next->next;cur->next->next = temp;cur->next->next->next = temp1;temp1 = temp->next;cur = temp;}return dummyhead->next;}
};

19.删除链表的倒数第N个节点

快慢指针，主要是思路

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* removeNthFromEnd(ListNode* head, int n) {ListNode *dummyhead = new ListNode(0);dummyhead ->next = head;ListNode *fast_point = dummyhead;ListNode *slow_point = dummyhead;n++;while(n--&&fast_point!=nullptr){fast_point = fast_point->next;}while(fast_point!=nullptr){fast_point = fast_point->next;slow_point = slow_point->next;}ListNode* temp = slow_point->next;slow_point->next = slow_point->next->next;delete temp;temp = nullptr;return dummyhead->next;}
};

链表相交

重点在于链表如果相交，后面的长度一定一样，然后再比较指针就行，leetcode的方法一看就贼高级贼简洁，但是没看懂，烦

/*** 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 *curA = headA;ListNode *curB = headB;int length_a = 0;int length_b = 0;while(curA!=nullptr){curA = curA->next;length_a++;}while(curB!=nullptr){curB = curB->next;length_b++;}curA = headA;curB = headB;if(length_b>length_a){swap(length_a,length_b);swap(curA,curB);}int gap = length_a-length_b;while(gap--){curA = curA->next;}while(curA!=nullptr){if(curA == curB){return curA;}curA = curA->next;curB = curB->next;}return nullptr;}
};

环形链表II

很有意思，得先想到fast走两步 slow走一步 ，然后就变成一道数学题了，学到了

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode(int x) : val(x), next(NULL) {}* };*/
class Solution {
public:ListNode *detectCycle(ListNode *head) {ListNode *fast = head;ListNode *slow = head;while(fast!=nullptr && fast->next !=nullptr){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 nullptr;}
};