[LeetCode] 021. Merge Two Sorted Lists (Easy) (C++/Python)

索引：[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql)

Github: https://github.com/illuz/leetcode

021.Merge_Two_Sorted_Lists (Easy)

链接：

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

代码(github)：https://github.com/illuz/leetcode

题意：

合并两个有序链表。

分析：

很经典的题目，不过知道怎么做后很容易，模拟即可。

有两种做法：

1. 开一个节点做 head 的前节点 （下面的 Python 代码实现）

2. 不开直接做（C++ 代码实现）

代码：

C++:

class Solution {
public:
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
		if (l1 == NULL)
			return l2;
		if (l2 == NULL)
			return l1;

		ListNode *start, *cur;

		if (l1->val < l2->val) {
			cur = start = l1;
			l1 = l1->next;
		} else {
			cur = start = l2;
			l2 = l2->next;
		}
		while (l1 != NULL && l2 != NULL) {
			if (l1->val < l2->val) {
				cur->next = l1;
				cur = l1;
				l1 = l1->next;
			} else {
				cur->next = l2;
				cur = l2;
				l2 = l2->next;
			}
		}
		if (l1 != NULL)
			cur->next = l1;
		else
			cur->next = l2;
		return start;
    }
};

ListNode *l1, *l2, *ll1, *ll2;
int main() {
	int n1, n2;
	Solution s;
	cin >> n1;
	ll1 = l1 = new ListNode(0);
	for (int i = 0; i < n1; i++) {
		l1->next = new ListNode(0);
		l1 = l1->next;
		scanf("%d", &(l1->val));
	}
	cin >> n2;
	ll2 = l2 = new ListNode(0);
	for (int i = 0; i < n2; i++) {
		l2->next = new ListNode(0);
		l2 = l2->next;
		scanf("%d", &(l2->val));
	}
	ListNode *res = s.mergeTwoLists(ll1->next, ll2->next);
	while (res != NULL) {
		cout << res->val << ' ';
		res = res->next;
	}

	return 0;
}

Python:

class Solution:
    # @param two ListNodes
    # @return a ListNode
    def mergeTwoLists(self, l1, l2):
        if not l1 and not l2:
            return None

        dummy = ListNode(0)
        cur = dummy
        while l1 and 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 or l2

        return dummy.next