DAY4 - 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

要画图，不然会乱。

class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        ListNode* fhead=new ListNode();
        fhead->next=head;
        ListNode* p=fhead;
        while(p->next&&p->next->next){
            ListNode* n1=p->next;
            ListNode* n2=p->next->next;
            p->next=n2;
            n1->next=n2->next;
            n2->next=n1;
            p=p->next->next;
        }
        ListNode* res=fhead->next;
        delete fhead;
        return res;
    }
};

19.删除链表的倒数第N个节点

这道题还有点印象，用双指针

一样的，画个图就清楚了。

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode* fhead=new ListNode();
        fhead->next=head;
        ListNode* last=fhead;
        ListNode* ln=fhead;
        for(int i=n;i>0;i--){
            last=last->next;
        }
        while(last->next){
            last=last->next;
            ln=ln->next;
        }
        ListNode* temp=ln->next;
        ln->next=temp->next;
        delete temp;

        return fhead->next;
    }
};

面试题02.07.链表相交

如果有交点，则尾端肯定对齐。因为不涉及增删操作，所以可以不用设置 fhead

（但我还是写了fhead）

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode* fheadA = new ListNode();
        ListNode* fheadB = new ListNode();
        fheadA->next = headA;
        fheadB->next = headB;

        int lenA = 0, lenB = 0;

        for (ListNode* a = fheadA; a != NULL; a = a->next) lenA++;
        for (ListNode* b = fheadB; b != NULL; b = b->next) lenB++;

        if (lenA == 0 || lenB == 0) return NULL;

        ListNode* a = headA;
        ListNode* b = headB;

        if (lenA > lenB) {
            for (; lenA > lenB; lenA--)  a = a->next;
        } else {
            for (; lenB > lenA; lenB--) b = b->next;
        }

        while (a != NULL && b != NULL) {
            if (a == b) return a;
            a = a->next;
            b = b->next;
        }

        return NULL;
    }
};

142.环形链表II