题目
给你一个链表数组,每个链表都已经按升序排列。
请你将所有链表合并到一个升序链表中,返回合并后的链表。
约束
解法
ListNode* mergeTwoLists(ListNode *a, ListNode *b) {
if ((!a) || (!b)) return a ? a : b;
ListNode head, *tail = &head, *aPtr = a, *bPtr = b;
while (aPtr && bPtr) {
if (aPtr->val < bPtr->val) {
tail->next = aPtr; aPtr = aPtr->next;
} else {
tail->next = bPtr; bPtr = bPtr->next;
}
tail = tail->next;
}
tail->next = (aPtr ? aPtr : bPtr);
return head.next;
}
解法一
class Solution {
public:
ListNode* mergeTwoLists(ListNode *a, ListNode *b) {
if ((!a) || (!b)) return a ? a : b;
ListNode head, *tail = &head, *aPtr = a, *bPtr = b;
while (aPtr && bPtr) {
if (aPtr->val < bPtr->val) {
tail->next = aPtr; aPtr = aPtr->next;
} else {
tail->next = bPtr; bPtr = bPtr->next;
}
tail = tail->next;
}
tail->next = (aPtr ? aPtr : bPtr);
return head.next;
}
ListNode* mergeKLists(vector<ListNode*>& lists) {
ListNode *ans = nullptr;
for (size_t i = 0; i < lists.size(); ++i) {
ans = mergeTwoLists(ans, lists[i]);
}
return ans;
}
};
解法二
class Solution {
public:
ListNode* mergeTwoLists(ListNode *a, ListNode *b) {
if ((!a) || (!b)) return a ? a : b;
ListNode head, *tail = &head, *aPtr = a, *bPtr = b;
while (aPtr && bPtr) {
if (aPtr->val < bPtr->val) {
tail->next = aPtr; aPtr = aPtr->next;
} else {
tail->next = bPtr; bPtr = bPtr->next;
}
tail = tail->next;
}
tail->next = (aPtr ? aPtr : bPtr);
return head.next;
}
ListNode* merge(vector <ListNode*> &lists, int l, int r) {
if (l == r) return lists[l];
if (l > r) return nullptr;
int mid = (l + r) >> 1;
return mergeTwoLists(merge(lists, l, mid), merge(lists, mid + 1, r));
}
ListNode* mergeKLists(vector<ListNode*>& lists) {
return merge(lists, 0, lists.size() - 1);
}
};
解法三
class Solution {
public:
struct Status {
int val;
ListNode *ptr;
bool operator < (const Status &rhs) const {
return val > rhs.val;
}
};
priority_queue <Status> q;
ListNode* mergeKLists(vector<ListNode*>& lists) {
for (auto node: lists) {
if (node) q.push({node->val, node});
}
ListNode head, *tail = &head;
while (!q.empty()) {
auto f = q.top(); q.pop();
tail->next = f.ptr;
tail = tail->next;
if (f.ptr->next) q.push({f.ptr->next->val, f.ptr->next});
}
return head.next;
}
};
标签:ListNode,next,链表,aPtr,tail,bPtr,升序,return,LeetCode
From: https://www.cnblogs.com/chuixulvcao/p/17090804.html