Leetcode 143. Reorder List

最新推荐文章于 2022-10-08 18:49:12 发布

原创最新推荐文章于 2022-10-08 18:49:12 发布 · 169 阅读

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本文介绍了一种链表重构算法，该算法将一个链表重新排序，使得原链表的前半部分与反转后的后半部分交错连接。文章详细阐述了算法的实现过程，包括寻找中间节点、反转链表后半部分以及交替合并两个链表。

文章作者：Tyan
博客：noahsnail.com | 优快云 | 简书

1. Description

Reorder List

2. Solution

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    void reorderList(ListNode* head) {
        if(!head) {
            return;
        }
        ListNode* p1 = head;
        ListNode* p2 = head;
        while(p2 && p2->next) {
            p1 = p1->next;
            p2 = p2->next->next;
        }
        ListNode* left = head;
        ListNode* right = reverseList(p1->next);
        p1->next = nullptr;
        ListNode* new_head = new ListNode(0);
        ListNode* current = new_head;
        while(left && right) {
            current->next = left;
            left = left->next;
            current = current->next;
            current->next = right;
            current = current->next;
            right = right->next;
        }
        current->next = left?left:right;
        head = new_head->next;
    }
    
private:
    ListNode* reverseList(ListNode* head) {
        ListNode* pre = nullptr;
        ListNode* next = nullptr;
        ListNode* current = head;
        while(current) {
            next = current->next;
            current->next = pre;
            pre = current;
            current = next;
        }
        return pre;
    }
};