LeetCode/排序链表

1. 合并两个有序链表

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode* preHead = new ListNode(-1);
        //一定要使用头指针并复制一个副本，不然后面循环操作不统一，未合并完的也不好加上去
        ListNode* prev = preHead;
        while (l1&&l2) {
            if (l1->val < l2->val) {
                prev->next = l1;//链入pre
                l1 = l1->next;//指针后移
            } else {
                prev->next = l2;
                l2 = l2->next;
            }
            prev = prev->next;//pre指针后移
        }
        // 合并后 l1 和 l2 最多只有一个还未被合并完，我们直接将链表末尾指向未合并完的链表即可
        prev->next = l1 == nullptr ? l2 : l1;

        return preHead->next;
    }
//也可以使用递归
};

2. 合并k个有序链表

//归并
ListNode* mergeKLists(vector<ListNode*>& lists) {
        if(!lists.size()) return {};
        return merge(lists,0,lists.size()-1);
    }

    ListNode* merge(vector<ListNode*>& lists,int i ,int j){
        if(i==j) return lists[i];//边界
        ListNode*left=merge(lists,i,(i+j)/2);//分治
        ListNode*right=merge(lists,(i+j)/2+1,j);//分治
        return mergeTwoLists(left,right);//归并
    }

//顺序合并
ListNode* mergeKLists(vector<ListNode*>& lists) {
        ListNode *ans = nullptr;
        for (size_t i = 0; i < lists.size(); ++i) {
            ans = mergeTwoLists(ans, lists[i]);
        }
        return ans;
    }

3. 排序链表(归并)

//主函数
ListNode* sortList(ListNode* head) {
        // 1、递归结束条件
        if (!head||!head->next)  return head;
        
        // 2、找到链表中间节点并断开链表(两个返回第一个)
        ListNode* midNode = middleNode(head);
        ListNode* rightHead = midNode->next;
        midNode->next = nullptr;

        ListNode* left = sortList(head);//分治
        ListNode* right = sortList(rightHead);//分治
        // 3、当前层业务操作（合并有序链表）
        return mergeTwoLists(left, right);//归并
    }

//找中间节点的前一个节点
ListNode* middleNode(ListNode* head) {
        if(!head||!head->next) return head;
        ListNode* slow = head;
        ListNode* fast = head->next->next;
        while (fast != NULL && fast->next != NULL) {
            slow = slow->next;
            fast = fast->next->next;
        }
        return slow;
    }

4. 排序链表(极简归并)

class Solution {
public:
    ListNode* sortList(ListNode* head) {
        if (!head || !head->next) return head;
        auto slow = head, fast = head;
        while (fast->next && fast->next->next)
            slow = slow->next, fast = fast->next->next;
        // 切链
        fast = slow->next, slow->next = nullptr;
        return merge(sortList(head), sortList(fast));
    }

private:
    ListNode* merge(ListNode* l1, ListNode* l2) {
        ListNode sub(0), *ptr = ⊂
        while (l1 && l2) {
            auto &node = l1->val < l2->val ? l1 : l2;
            ptr = ptr->next = node, node = node->next;
        }
        ptr->next = l1 ? l1 : l2;
        return sub.next;
    }
};

标签：head,ListNode,next,链表,l2,l1,return,排序,LeetCode
来源： https://www.cnblogs.com/929code/p/16518138.html