【LeetCode】C++ ：中等题 - 链表 143. 重排链表

143. 重排链表

难度中等511

给定一个单链表 L：L0→L1→…→Ln-1→Ln ，
将其重新排列后变为： L0→Ln→L1→Ln-1→L2→Ln-2→…

你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

示例 1:

给定链表 1->2->3->4, 重新排列为 1->4->2->3.

示例 2:

给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.

1、链表数组存储链表，然后利用数组的索引进行重新排列

/**
 * 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:
    void reorderList(ListNode* head) {
        if(head == nullptr){
            return ;
        }
        vector<ListNode*> vec;
        ListNode* node = head;
        while(node != nullptr){
            vec.push_back(node);
            node = node->next;
        }
        int i = 0, j = vec.size()-1;
        while(i < j){
            vec[i]->next = vec[j];
            i++;
            if(i == j){
                break;
            }
            vec[j]->next = vec[i];
            j--;
        }
        vec[i]->next = nullptr;
    }
};

2、找中点-反转-合并

大概的思路是这样的：

先找到链表的中点，然后对中点之后的节点全部反转，最后合并的过程比较复杂需要仔细的在纸上画画过程找找规律。总之，在官方题解中学习到很多。

/**
 * 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:
    void reorderList(ListNode* head) {
        if(head == nullptr){
            return ;
        }
        ListNode* mid = middleNode(head);
        ListNode* l1 = head;
        ListNode* l2 = mid->next;
        mid->next = nullptr;
        l2 = reverseList(l2);
        mergeList(l1, l2);
        
    }
    ListNode* middleNode(ListNode* head){
        ListNode* slow = head;
        ListNode* fast = head;
        while(fast->next != nullptr && fast->next->next != nullptr) {
            slow = slow->next;
            fast = fast->next->next;
        }
        return slow;
    }

    ListNode* reverseList(ListNode* head){
        ListNode* prev = nullptr;
        ListNode* curr = head;
        while(curr != nullptr){
            ListNode* tmp = curr->next;
            curr->next = prev;
            prev = curr;
            curr = tmp;
        }
        return prev;
    }

    void mergeList(ListNode* l1, ListNode* l2){
        ListNode* l1_tmp, *l2_tmp;
        while(l1 != nullptr && l2 != nullptr){
            l1_tmp = l1->next;
            l2_tmp = l2->next;
            l1->next = l2;
            l1 = l1_tmp;
            l2->next = l1;
            l2 = l2_tmp;
        }
    }
};

/**
 * 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:
    void reorderList(ListNode* head) {
        if(head == nullptr || head->next == nullptr || head->next->next == nullptr){
            return;
        }
        ListNode *slow = head, *fast = head->next;
        while(fast != nullptr && fast->next != nullptr){
            slow = slow->next;
            fast = fast->next->next;
        }   //找中点

        //2 断开中点后，反转后半部分
        ListNode *head2 = nullptr, *nxt = slow->next;
        slow->next = nullptr;
        slow = nxt;
        while(slow != nullptr){
            nxt = slow->next;
            slow->next = head2;
            head2 = slow;
            slow = nxt;
        }

        //3 合并链表
        ListNode* curr = head, *curr2 = head2;
        while(curr != nullptr && curr2 != nullptr){
            nxt = curr->next;
            curr->next = curr2;
            curr2 = curr2->next;
            curr->next->next = nxt;
            curr = nxt;
        }

    }
};