代码随想录刷题第4天|LeetCode24、LeetCode19、LeetCode160、LeetCode142

1、LeetCode24 两两交换链表中的节点

题目链接：24、两两交换链表中的节点

要想清楚终止条件，cur每次指向要交换的两个节点的前一个节点，cur = cur->next->next;

若链表元素个数为偶数 , 则最后时刻 cur->next = NULL；

若链表元素个数为奇数，则最后时刻 cur->next->next = NULL;

最后要返回 dummyHead->next, 因为交换节点后 head 为 “2”，不是原来的 “1”。

class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        ListNode * dummyHead = new ListNode(0);
        dummyHead->next = head;
        ListNode * cur = dummyHead;

        while(cur->next!=NULL && cur->next->next!=NULL)
        {
            ListNode * temp = cur->next;
            ListNode * temp1 = cur->next->next->next;

            cur->next = cur->next->next;
            cur->next->next = temp;
            cur->next->next->next = temp1;

            cur = cur->next->next;
        }

        return dummyHead->next;
    }
};

2、LeetCode19 删除链表的倒数第N个节点

题目链接：19、删除链表的倒数第N个节点

用快慢指针，快指针先向后移动N+1个节点，然后当快指针不为NULL时，快慢指针一起向后移动，快指针为NULL时，慢指针刚好移动到 倒数第N个节点的前一个节点。

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode * dummyHead = new ListNode(0);
        dummyHead->next = head;
        ListNode * fast = dummyHead;
        ListNode * slow = dummyHead;

        n++;
        while(n-- && fast!=NULL)
        {
            fast = fast->next;
        }
        while(fast!=NULL)
        {
            fast = fast->next;
            slow = slow->next;
        }

        ListNode * temp = slow->next;
        slow->next = slow->next->next;
        delete temp;

        return dummyHead->next;
    }
};

3、LeetCode160 链表相交

题目链接：160 链表相交

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode * cura = headA;
        ListNode * curb = headB;
        int lenA = 0;
        int lenB = 0;

        while(cura)
        {
            lenA++;
            cura = cura->next;
        }
        while(curb)
        {
            lenB++;
            curb = curb->next;
        }
        cura = headA;
        curb = headB;
        
        // 让curA为最长链表的头，lenA为其长度
        if (lenB > lenA)
        {
            swap(lenA,lenB);
            swap(cura,curb);
        }

        //求长度差
        int gap = lenA - lenB;
        while (gap--)
        {
            cura = cura->next;
        }

        while(cura)
        {
            if (cura == curb)
            {
                return cura;
            }
            cura = cura->next;
            curb = curb->next;
        }
        return NULL;
    }
};

4、LeetCode142 环形链表

题目链接：142、环形链表

找相遇节点，相遇节点和头结点同时出发，相遇时即为环形入口节点。（太厉害了，自己是想不出来。）

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        ListNode * fast = head;
        ListNode * slow = head;

        while (fast!=NULL && fast->next!=NULL)
        {

            slow = slow->next;
            fast = fast->next->next;
            if(slow == fast)
            {
                ListNode * index1 = head;
                ListNode * index2 = fast;
                while (index1 != index2)
                {
                    index1 = index1->next;
                    index2 = index2->next;
                }
                return index1;
            }
        }
        return NULL;
    }
};