代码随想录算法训练营第四天 | 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

/**
 * 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 && cur->next->next){
            ListNode *tmp = cur->next;
            ListNode *tmp1 = cur->next->next->next;
            cur->next = tmp->next;
            cur->next->next = tmp;
            cur->next->next->next = tmp1;
            cur = tmp;
        }
        return dummyhead->next;
    }
};

心得：还是对链表的细节理解不透彻。

19. 删除链表的倒数第 N 个结点

/**
 * 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 *slow = dummyhead;;
        ListNode *fast = dummyhead;
        while(n-- && fast){
            fast = fast->next;
        }
        fast = fast->next;
        while(fast){
            slow = slow->next;
            fast = fast->next;
        }
        slow->next = slow->next->next;
        return dummyhead->next;
    }
};

心得：最后不能直接返回head，要考虑链表中只有一个节点且删除倒数第一个节点的情况。

面试题 02.07. 链表相交

/**
 * 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 lenA = 0, lenB = 0;
        while(curA){
            lenA++;
            curA = curA->next;
        }
        while(curB){
            lenB++;
            curB = curB->next;
        }
        curA = headA;
        curB = headB;
        if(lenA < lenB){
            swap(curA, curB);
            swap(lenA, lenB);
        }
        int gap = lenA - lenB;
        while(gap--){
            curA = curA->next;
        }
        while(curA && curB){
            if(curA == curB) return curB;
            else{
                curA = curA->next;
                curB = curB->next;
            }
        }
        return nullptr;
    }
};

心得：

1. 这题的思路确实想不出来，没有想到要将A与B链表末尾对齐。
2. 要看清楚函数的返回值。

142. 环形链表 II

/**
 * 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 *slow = head;
        ListNode *fast = head;
        while(fast && fast->next){
            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;
    }
};

心得：这题确实想不出来，中间很多推理都是我做不到的。