Merge k Sorted Lists
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
分析
The basic idea is really simple. We can merge first two lists and then push it back. Keep doing this until there is only one list left in vector. Actually, we can regard this as an iterative divide-and-conquer solution.
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 *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;
}
}
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原文地址:http://blog.csdn.net/lmingyin5/article/details/47777755