LeetCode 147. Insertion Sort List

最新推荐文章于 2024-11-10 21:15:35 发布

原创最新推荐文章于 2024-11-10 21:15:35 发布 · 251 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#LeetCode 147. Insertion Sort L

LeetCode练习题专栏收录该内容

168 篇文章

订阅专栏

本文提供了两种实现LeetCode 147题“插入排序链表”的解决方案。第一种方法较为直观但效率较低；第二种方法借鉴他人思路，代码更为简洁高效。通过对两种方法的对比，可以学习到如何优化链表操作。

LeetCode 147. Insertion Sort List

Solution1：我的答案
有点笨，有点慢

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* insertionSortList(ListNode* head) {
        if (!head) return NULL;
        if (!head->next) return head;
        ListNode* new_head = new ListNode(-1), *cur = head;
        while (cur) {
            ListNode *temp = cur->next;
            cur->next = NULL;
            my_insert(new_head, cur);
            cur = temp;
        }
        return new_head->next;
    }

    void my_insert (ListNode* &new_head, ListNode* &des) {
        if (!new_head->next || des->val <= new_head->next->val) {
            ListNode *temp = new_head->next;
            new_head->next = des;
            des->next = temp;
            return;
        } else {
            ListNode* cur = new_head->next;
            while (cur->next) {
                if (des->val >= cur->val && des->val <= cur->next->val) {
                    des->next = cur->next;
                    cur->next = des;
                    return;
                }
                else
                    cur = cur->next;
            }
            cur->next = des;
            return;
        }
    }
};

Solution2：
参考网址：http://www.cnblogs.com/grandyang/p/4250107.html

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* insertionSortList(ListNode* head) {
        ListNode *dummy = new ListNode(-1), *cur = dummy;
        while (head) {
            ListNode *t = head->next;//暂时保存头结点的下一个位置
            cur = dummy;
            while (cur->next && cur->next->val <= head->val) {
                cur = cur->next;
            }
            head->next = cur->next;
            cur->next = head;
            head = t;
        }
        return dummy->next;
    }
};