LeetCode 21.合并两个有序链表（归并思想）

题目描述

原题链接

算法

(归并思想) $O(n+m)$

• 分别比较两个链表大小，然后用归并的思想连接这些点即可

时间复杂度是 $O(n + m)$，空间复杂度是 $O(1)$

代码

非递归写法

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        auto head = new ListNode(-1);
        auto p = head;
        while(l1 && l2) {
            if (l1->val <= l2->val) {
                p->next = l1;
                p = l1;
                l1 = l1->next;
            } else {
                p->next = l2;
                p = l2;
                l2 = l2->next;
            }
        }

        // There will be a line of code is executed
        if(l1) p->next = l1;
        if(l2) p->next = l2;

        return head->next;
    }
};

递归写法

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        // Recursion
        // 1 2 5 NULL
        //   3 4 7 NULL    
        if (l1 == nullptr) return l2;
        if (l2 == nullptr) return l1;

        if (l1->val < l2->val) {
            l1->next = mergeTwoLists(l1->next, l2);
            return l1;
        }
        else {
            l2->next = mergeTwoLists(l1, l2->next);
            return l2;
        }
        
        return nullptr;
    }
};

补充：两个有序递增链表合并成一个递减链表（头插法）

一般常用的插入方法有：尾插法和头插法。这里使用头插法

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        auto End = new ListNode(-1);
        auto p = End->next; // p == NULL
        while(l1 && l2) {
            if (l1->val <= l2->val) {
                // 头插法
                auto tmp = l1->next;
                l1->next = p;
                p = l1;
                l1 = tmp;
            } else {
                auto tmp = l2->next;
                l2->next = p;
                p = l2;
                l2 = tmp;
            }
        }

        while(l1) {
            auto tmp = l1->next;
            l1->next = p;
            p = l1;
            l1 = tmp;
        }
        while(l2) {
            auto tmp = l2->next;
            l2->next = p;
            p = l2;
            l2 = tmp;
        }

        return p;
    }
};