[LeetCode] 21. Merge Two Sorted Lists 混合插入有序鏈表

 

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

 

這道混合插入有序鏈表和我之前那篇混合插入有序數組非常的相似 Merge Sorted Array，僅僅是數據結構由數組換成了鏈表而已，代碼寫起來反而更簡潔。具體思想就是新建一個鏈表，然后比較兩個鏈表中的元素值，把較小的那個鏈到新鏈表中，由於兩個輸入鏈表的長度可能不同，所以最終會有一個鏈表先完成插入所有元素，則直接另一個未完成的鏈表直接鏈入新鏈表的末尾。代碼如下：

 

C++ 解法一：

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode *dummy = new ListNode(-1), *cur = dummy;
        while (l1 && l2) {
            if (l1->val < l2->val) {
                cur->next = l1;
                l1 = l1->next;
            } else {
                cur->next = l2;
                l2 = l2->next;
            }
            cur = cur->next;
        }
        cur->next = l1 ? l1 : l2;
        return dummy->next;
    }
};

 

Java 解法一：

public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(-1), cur = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                cur.next = l1;
                l1 = l1.next;
            } else {
                cur.next = l2;
                l2 = l2.next;
            }
            cur = cur.next;
        }
        cur.next = (l1 != null) ? l1 : l2;
        return dummy.next;
    }
}

 

下面我們來看遞歸的寫法，當某個鏈表為空了，就返回另一個。然后核心還是比較當前兩個節點值大小，如果 l1 的小，那么對於 l1 的下一個節點和 l2 調用遞歸函數，將返回值賦值給 l1.next，然后返回 l1；否則就對於 l2 的下一個節點和 l1 調用遞歸函數，將返回值賦值給 l2.next，然后返回 l2，參見代碼如下：

 

C++ 解法二：

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if (!l1) return l2;
        if (!l2) return l1;
        if (l1->val < l2->val) {
            l1->next = mergeTwoLists(l1->next, l2);
            return l1;
        } else {
            l2->next = mergeTwoLists(l1, l2->next);
            return l2;
        }
    }
};

 

Java 解法二：

public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        if (l1.val < l2.val) {
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        } else {
            l2.next = mergeTwoLists(l1, l2.next);
            return l2;
        }
    }
}

 

下面這種遞歸的寫法去掉了 if 從句，看起來更加簡潔一些，但是思路並沒有什么不同：

 

C++ 解法三：

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if (!l1) return l2;
        if (!l2) return l1;
        ListNode *head = l1->val < l2->val ? l1 : l2;
        ListNode *nonhead = l1->val < l2->val ? l2 : l1;
        head->next = mergeTwoLists(head->next, nonhead);
        return head;
    }
};

 

Java 解法三：

public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        ListNode head = (l1.val < l2.val) ? l1 : l2;
        ListNode nonhead = (l1.val < l2.val) ? l2 : l1;
        head.next = mergeTwoLists(head.next, nonhead);
        return head;
    }
}

 

 我們還可以三行搞定，簡直喪心病狂有木有！

 

C++ 解法四：

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if (!l1 || (l2 && l1->val > l2->val)) swap(l1, l2);
        if (l1) l1->next = mergeTwoLists(l1->next, l2);
        return l1;
    }
};

 

Java 解法四：

public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null || (l2 != null && l1.val > l2.val)) {
            ListNode t = l1; l1 = l2; l2 = t;
        }
        if (l1 != null) l1.next = mergeTwoLists(l1.next, l2);
        return l1;
    }
}

 

Github 同步地址：

https://github.com/grandyang/leetcode/issues/21

 

類似題目：

Merge Sorted Array

Merge k Sorted Lists

Sort List

Shortest Word Distance II 

 

參考資料：

https://leetcode.com/problems/merge-two-sorted-lists/

https://leetcode.com/problems/merge-two-sorted-lists/discuss/9714/14-line-clean-C%2B%2B-Solution

https://leetcode.com/problems/merge-two-sorted-lists/discuss/9814/3-lines-C%2B%2B-(12ms)-and-C-(4ms)

https://leetcode.com/problems/merge-two-sorted-lists/discuss/9715/Java-1-ms-4-lines-codes-using-recursion

 

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