合并K个有序链表----链表OJ

https://leetcode.cn/problems/merge-k-sorted-lists/submissions/499384099/?envType=study-plan-v2&envId=top-100-liked

        1、两两合并

        前面我们做过合并两个有序链表,那么这里合并K个有序链表,是否可以联想到合并两个呢?答案是可以的!两两合并

        思路:

        1、将lists[0] 和 lists[1]合并,返回给lists[0],此时lists[0]就是0,1合并后的链表。

        2、将lists[0] 和 lists[2]合并,返回给lists[0],此时lists[0]就是0,1,2合并后的链表。

        ........

        最后、将lists[0]和lists[K-1]合并,返回给lists[0],那么此时lists[0]就是最终合并后的链表。

        则最终代码:

//两两合并
struct ListNode* merge(struct ListNode* head1, struct ListNode* head2) {
    struct ListNode* dummy = (struct ListNode*)malloc(sizeof(struct ListNode));
    dummy->val = 0, dummy->next = NULL;
    struct ListNode *tmp = dummy, *h1 = head1, *h2 = head2;
    while (h1 && h2) {
        if (h1->val <= h2->val) {
            tmp->next = h1;
            h1 = h1->next;
        } else {
            tmp->next = h2;
            h2 = h2->next;
        }
        tmp = tmp->next;
    }
    if (h1) {
        tmp->next = h1;
    }
    if (h2) {
        tmp->next = h2;
    }
    return dummy->next;
}

struct ListNode* mergeKLists(struct ListNode** lists, int listsSize) {
    if(lists == NULL)//判空
        return NULL;
    if(listsSize == 0)//判断是否有K == 0的情况
        return NULL;
    for(int i = 1;i

        2、整合排序

        前面我们也写过,将一个乱序的链表排序,那么这里是否可以先将所有链表连接成为一个新链表,再对这个新链表排序呢?答案是可以!

        连接链表代码如下:

    struct ListNode* dummyhead = malloc(sizeof(struct ListNode));
    dummyhead->val = 0,dummyhead->next = NULL;
    struct ListNode* curr = dummyhead;
    for(int i = 0;inext = lists[i];
        while(curr->next)
        {
            curr = curr->next;
        }
    }
    struct ListNode* head = dummyhead->next;

        用head来代表这个新链表的头。之后我们将前面写的链表排序代码copy一下即可。总代码如下:

struct ListNode* mergeKLists(struct ListNode** lists, int listsSize) {
    struct ListNode* dummyhead = malloc(sizeof(struct ListNode));
    dummyhead->val = 0,dummyhead->next = NULL;
    struct ListNode* curr = dummyhead;
    for(int i = 0;inext = lists[i];
        while(curr->next)
        {
            curr = curr->next;
        }
    }
    struct ListNode* head = dummyhead->next;

    int len = 0; //长度
    for (struct ListNode* cur = head; cur != NULL; cur = cur->next)
        len++;

    struct ListNode* dummy = malloc(sizeof(struct ListNode));
    dummy->val = 0, dummy->next = head;

    //自底向上归并排序
    for (int sublen = 1; sublen < len; sublen *= 2) {
        struct ListNode *pre = dummy, *cur = dummy->next; //每次从新的头开始记录
        while (cur) {
            struct ListNode* head1 = cur; //第一个头就是cur
            for (int i = 1; i < sublen && cur->next != NULL;
                 i++) //找1子区间的尾,并且2子区间不为空
            {
                cur = cur->next;
            }
            //如果for是在cur->next == NULL结束的,那2子区间头就是空
            struct ListNode* head2 = cur->next; // 2子区间的头
            cur->next = NULL;                   //将1子区间分离出来
            cur = head2;
            //再找2子区间的尾
            for (int i = 1; i < sublen && cur != NULL; i++) {
                cur = cur->next;
            }

            struct ListNode* next = NULL; //记录下一组的头
            //如果cur为空,说明已经到了整个链表的最后
            if (cur != NULL) // cur不为空
            {
                next = cur->next; //记录下一组的头,可空可不空
                cur->next = NULL; //分离2子区间
            }
            struct ListNode* Merged = merge(head1, head2); //记录每次合并后的头
            pre->next = Merged;
            while (pre->next) //走到合并后的1,2区间的尾,pre来链接每一组
            {
                pre = pre->next;
            }
            cur = next; //进入下一组
        }
    }
    return dummy->next;
}

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