题解 | #合并k个已排序的链表#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param lists ListNode类一维数组 
 * @param listsLen int lists数组长度
 * @return ListNode类
 */
 #include <stdlib.h>
 //合并两个链表
struct ListNode*Linklist(struct ListNode*head1,struct ListNode*head2){
    if(head1==NULL)return head2;
    if(head2==NULL)return head1;
    struct ListNode*H = (struct ListNode*)malloc(sizeof(struct ListNode));
    struct ListNode*cur = H;
    while(head1 && head2){
        if(head1->val<=head2->val){
            cur->next=head1;
            head1=head1->next;
            cur=cur->next;
        }else {
            cur->next=head2;
            head2=head2->next;
            cur=cur->next;
        }
        if(head1==NULL){
            cur->next=head2;
        }
        if(head2==NULL){
            cur->next=head1;
        }
    }
    return H->next;
 }
struct ListNode* mergeKLists(struct ListNode** lists, int listsLen ) {
    // write code here
    if(listsLen==0 || listsLen ==1)return *lists;
    struct ListNode*r = lists[0];
    for(int i = 1;i< listsLen;i++){
        //从第0个链表和第1个链表合并开始，两两合并成新链表再和下一个链表合并，一直循环到与最后一个链表合并
        r = Linklist(r,lists[i]);
    }
        
    return r;


}