• [LeetCode] 21. Merge Two Sorted Lists 混合插入有序链表


    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

    这道混合插入有序链表和我之前那篇混合插入有序数组非常的相似 Merge Sorted Array,仅仅是数据结构由数组换成了链表而已,代码写起来反而更简洁。具体思想就是新建一个链表,然后比较两个链表中的元素值,把较小的那个链到新链表中,由于两个输入链表的长度可能不同,所以最终会有一个链表先完成插入所有元素,则直接另一个未完成的链表直接链入新链表的末尾。代码如下:

    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;
        }
    }

    下面我们来看递归的写法,当某个链表为空了,就返回另一个。然后核心还是比较当前两个节点值大小,如果 l1 的小,那么对于 l1 的下一个节点和 l2 调用递归函数,将返回值赋值给 l1.next,然后返回 l1;否则就对于 l2 的下一个节点和 l1 调用递归函数,将返回值赋值给 l2.next,然后返回 l2,参见代码如下:

    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;
            }
        }
    }

    下面这种递归的写法去掉了 if 从句,看起来更加简洁一些,但是思路并没有什么不同:

    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;
        }
    }

     我们还可以三行搞定,简直丧心病狂有木有!

    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

    类似题目:

    Merge Sorted Array

    Merge k Sorted Lists

    Sort List

    Shortest Word Distance II 

    参考资料:

    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

    LeetCode All in One 题目讲解汇总(持续更新中...)

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  • 原文地址:https://www.cnblogs.com/grandyang/p/4086297.html
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