206.反转链表
#leetcode#链表 2025-01-26

Problem: 206. 反转链表

思路

这个题干想很费脑子，多个指针变来变去可记不住，浪费一个小时也没调对，还是得画下图，一步一步规划代码，类似数形结合思想，这叫码形结合思想

解题过程

开头额外处理，剩下的节点循环处理即可

复杂度

  • 时间复杂度: $O(n)$ n次处理指针过程
  • 空间复杂度: $O(1)$

Code

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//! 码形结合思想 n,1
class Solution {
public:
  ListNode *reverseList(ListNode *head) {
    if (head == nullptr) {
      return nullptr;
    }

    ListNode *node = head;
    ListNode *pre;
    ListNode *nex;
    pre = node;
    if (node->next == nullptr) {
      return node;
    }
    node = node->next;
    nex = node->next;
    node->next = pre;
    pre->next = nullptr;
    pre = node;
    node = nex;
    while (node != nullptr) {
      nex = node->next;
      node->next = pre;
      pre = node;
      node = nex;
    }
    return pre;
  }
};
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//! 码形结合思想 n,1 上个答案对不必要逻辑和变量的缩减
class Solution {
public:
  ListNode *reverseList(ListNode *head) {

    ListNode *node = head;
    ListNode *pre = nullptr;
    ListNode *nex;
    pre = node;
    if (node->next == nullptr) {
      return node;
    }
    node = node->next;
    nex = node->next;
    node->next = pre;
    pre->next = nullptr;
    pre = node;
    node = nex;
    while (node != nullptr) {
      nex = node->next;
      node->next = pre;
      pre = node;
      node = nex;
    }
    return pre;
  }
};
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//! 错误答案，第一遍给干成循环移动链表的，好好审题，浪费时间了
class Solution {
public:
  ListNode *reverseList(ListNode *head) {
    if (head == nullptr) {
      return nullptr;
    }

    ListNode *node1 = head;

    ListNode *tail = head;
    while (tail->next != nullptr) {
      tail = tail->next;
    }
    ListNode *node2 = tail;

    while (node1->next != tail) {
      node2->next = node1;
      node2 = node2->next;
      node1 = node1->next;
    }
    node2->next = node1;
    node1->next = nullptr; // 破坏循环链表

    return tail;
  }
};