题目
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
频度： 4
解题之法
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
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()); // 注意vector的erase 方法
}
return lists.front();
}
ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
if(l1 == nullptr){
return l2;
}
if(l2 == nullptr){
return l1;
}
if(l1->val <= l2->val){
l1->next = mergeTwoLists(l1->next, l2);
return l1;
}
else{
l2->next = mergeTwoLists(l1, l2->next);
return l2;
}
}
};
分析
与合并2个链表的解法类似，只不过迭代地做。
