sorting/merge-sort/merge-k-sorted-lists
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2023-12-01
Merge k Sorted Lists
描述
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
分析
可以复用 Merge Two Sorted Lists 的函数
代码
// Merge k Sorted Lists
// 时间复杂度O(n1+n2+...),空间复杂度O(1)
// TODO: 会超时
public class Solution {
public ListNode mergeKLists(ListNode[] lists) {
if (lists.length == 0) return null;
ListNode p = lists[0];
for (int i = 1; i < lists.length; i++) {
p = mergeTwoLists(p, lists[i]);
}
return p;
}
// Merge Two Sorted Lists
ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode head = new ListNode(-1);
for (ListNode p = head; l1 != null || l2 != null; p = p.next) {
int val1 = l1 == null ? Integer.MAX_VALUE : l1.val;
int val2 = l2 == null ? Integer.MAX_VALUE : l2.val;
if (val1 <= val2) {
p.next = l1;
l1 = l1.next;
} else {
p.next = l2;
l2 = l2.next;
}
}
return head.next;
}
}
// Merge k Sorted Lists
// 时间复杂度O(n1+n2+...),空间复杂度O(1)
// TODO: 会超时
class Solution {
public:
ListNode *mergeKLists(vector<ListNode *> &lists) {
if (lists.size() == 0) return nullptr;
ListNode *p = lists[0];
for (int i = 1; i < lists.size(); i++) {
p = mergeTwoLists(p, lists[i]);
}
return p;
}
// Merge Two Sorted Lists
ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
ListNode head(-1);
for (ListNode* p = &head; l1 != nullptr || l2 != nullptr; p = p->next) {
int val1 = l1 == nullptr ? INT_MAX : l1->val;
int val2 = l2 == nullptr ? INT_MAX : l2->val;
if (val1 <= val2) {
p->next = l1;
l1 = l1->next;
} else {
p->next = l2;
l2 = l2->next;
}
}
return head.next;
}
};