Merge Two Sorted Lists Leetcode

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.

思想和merge sort一样，只是要考虑相等的情况的时候该怎么办。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        }
        if (l2 == null) {
            return l1;
        }
        ListNode result = new ListNode(0);
        ListNode returnResult = result;
        while (l1 != null && l2 != null) {
            while (l1 != null && l2 != null && l1.val < l2.val) {
                result.next = l1;
                l1 = l1.next;
                result = result.next;
            }
            while (l1 != null && l2 != null && l1.val > l2.val) {
                result.next = l2;
                l2 = l2.next;
                result = result.next;
            }
            while (l1 != null && l2 != null && l1.val == l2.val) {
                result.next = l1;
                l1 = l1.next;
                result = result.next;
                result.next = l2;
                l2 = l2.next;
                result = result.next;
            }
            
        }
        if (l1 == null) {
            result.next = l2;
        } else {
            result.next = l1;
        }
        return returnResult.next;
    }
}

写的还是有些冗余了。。。

简洁版：

public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        }
        if (l2 == null) {
            return l1;
        }
        ListNode result = new ListNode(0);
        ListNode returnResult = result;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                result.next = l1;
                l1 = l1.next;
            } else {
                result.next = l2;
                l2 = l2.next;
            }
            result = result.next;
        }
        if (l1 == null) {
            result.next = l2;
        } else {
            result.next = l1;
        }
        return returnResult.next;
    }
}

没想到还可以用recursion做

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

 

时隔两个月再来做这道题，发现自己还是进步了的~至少直接写出了简洁版。吼吼吼

也又写了一遍recursion

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) {
            l2.next = mergeTwoLists(l1, l2.next);
            return l2;
        } else {
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        }
    }
}

 

C++版本：

/**
 * 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) {
        if (l1 == NULL) {
            return l2;
        }
        if (l2 == NULL) {
            return l1;
        }
        if (l1 -> val > l2 -> val) {
            l2 -> next = mergeTwoLists(l1, l2 -> next);
            return l2;
        } else {
            l1 -> next = mergeTwoLists(l1 -> next, l2);
            return l1;
        }
    }
};

 c++另一种写法怎么也写不过。。。不懂指针什么的，到时候懂了再来写吧。。。#未完待续#需回顾。。。