c++学习笔记：有序链表的合并

/*
LeetCode_0023
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
*/
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};

1. 以递归的方式

ListNode* mergeTwo(ListNode* l1, ListNode* l2) {
if (l1 == NULL)
return l2;
if (l2 == NULL)
return l1;
if (l1->val <= l2->val) {
l1->next = mergeTwo(l1->next, l2);
return l1;
}
if (l1->val > l2->val) {
l2->next = mergeTwo(l1, l2->next);
return l2;
}
}

2. 直接合并

ListNode* mergeTwo(ListNode* l1, ListNode* l2) {
ListNode dummy(-1);
ListNode* result = &dummy;
while (l1&&l2) {
if (l1->val <= l2->val) {
result->next = l1;
result = result->next;
l1 = l1->next;
}
else if (l1->val > l2->val) {
result->next = l2;
result = result->next;
l2 = l2->next;
}
}
while (l1)
result->next = l1;
while (l2)
result->next = l2;
return dummy.next;
}

3. 合并k个有序链表

ListNode *mergeKLists(vector<ListNode *> &lists) {
if (lists.empty()) {
return nullptr;
}
while (lists.size() > 1) {
lists.push_back(mergeTwoLists(lists[0], lists[1]));
lists.erase(lists.begin());
lists.erase(lists.begin());
}
return lists.front();
}
ListNode *mergeKLists(vector<ListNode *> &lists) {
if (lists.empty()) return NULL;
int len = lists.size();
while (len > 1)
{
for (int i = 0; i < len / 2; ++i)
{
lists[i] = mergeTwoLists(lists[i], lists[len - 1 - i]);
}
len = (len + 1) / 2;
}
return lists.front();
}