LeetCode 24. Swap Nodes in Pairs

最新推荐文章于 2024-07-26 10:48:09 发布

原创最新推荐文章于 2024-07-26 10:48:09 发布 · 252 阅读

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本文介绍了一种链表中成对交换节点的算法实现，包括迭代和递归两种方式。迭代方法通过自定义反转部分链表的方法来完成节点交换，而递归方法则直接在递归函数内部进行节点交换，提供了简洁的解决方案。

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I did very bad on this one.....

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverse(ListNode* pre, ListNode* end){
        ListNode* last = pre->next;
        ListNode* cur = last->next;
        while(cur != end) {
            last->next = cur->next;
            cur->next = pre->next;
            pre->next = cur;
            
            cur = last->next;
        }
        return last;
    }
    ListNode *swapPairs(ListNode *head) {
        ListNode dummy(0);
        dummy.next = head;
        ListNode* pre = &dummy;
        int i = 0;
        while(head) {
            if(++i % 2 == 0) {
                pre = reverse(pre, head->next);
                head = pre->next;
            } else {
                head = head->next;
            }
        }
        return dummy.next;
    }
};

Do it recursively

  ListNode* swapNodesInRecursion(ListNode* head) {
    if(!head || !head->next) return head;
    ListNode* prev = head;
    ListNode* last = head->next;
    ListNode* tmp = swapNodesInRecursion(last->next);
    last->next = prev;
    prev->next = tmp;
    return last;
  }