23. 合併K個排序鏈表
合併 k 個排序鏈表,返回合併後的排序鏈表。請分析和描述算法的複雜度。
示例:
輸入: [ 1->4->5, 1->3->4, 2->6 ] 輸出: 1->1->2->3->4->4->5->6
雖然說是一道困難題 但是真的很簡單,
方法一:用vector<int>保存數據 然後排序 然後生成鏈表就好啦
/**
* 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) {
ListNode *dummy = new ListNode(-1), *cur = dummy;
vector<int> v1({});
for (auto node : lists) {
ListNode *t = node;
while (t) {
v1.push_back(t->val);
t = t->next;
}
}
sort(v1.begin(),v1.end());
for (int i = 0; i < v1.size(); ++i) {
cur->next = new ListNode(v1[i]);
cur = cur->next;
}
return dummy->next;
}
};
從別人的基數排序那裏獲得的啓發
方法二:基數排序
也學習一下方法
class Solution {
public:
ListNode* mergeKLists(vector<ListNode*>& lists) {
ListNode *dummy = new ListNode(-1), *cur = dummy;
unordered_map<int, int> m;
int mx = INT_MIN, mn = INT_MAX;
for (auto node : lists) {
ListNode *t = node;
while (t) {
mx = max(mx, t->val);
mn = min(mn, t->val);
++m[t->val];
t = t->next;
}
}
for (int i = mn; i <= mx; ++i) {
if (!m.count(i)) continue;
for (int j = 0; j < m[i]; ++j) {
cur->next = new ListNode(i);
cur = cur->next;
}
}
return dummy->next;
}
};
方法三:分治 最後兩兩對比 (重要!)
其中有些細節需要注意 k 和 n 的更新 等等
class Solution {
public:
ListNode* mergeKLists(vector<ListNode*>& lists) {
if (lists.empty()) return NULL;
int n = lists.size();
while (n > 1) {
int k = (n + 1) / 2;
for (int i = 0; i < n / 2; ++i) {
lists[i] = mergeTwoLists(lists[i], lists[i + k]);
}
n = k;
}
return lists[0];
}
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode *dummy = new ListNode(-1), *cur = dummy;
while (l1 && l2) {
if (l1->val < l2->val) {
cur->next = l1;
l1 = l1->next;
} else {
cur->next = l2;
l2 = l2->next;
}
cur = cur->next;
}
if (l1) cur->next = l1;
if (l2) cur->next = l2;
return dummy->next;
}
};