LeetCode-23-合并K个排序链表-C语言

方法一：递归分治
使用递归的方法，将数组不断分下去，一分为二，
递归截至条件为，数组中元素为0个，或者1个。
对分治后的两个链表进行合并排序。
分的时间复杂度为O（lgN），合并的时间复杂度为O(N),因此时间复杂度为O(NlgN);

typedef struct ListNode Node;

struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode head;
    head.next = NULL;
    struct ListNode *rear = &head;
    struct ListNode *node = NULL;
    
    while(l1 && l2){
        rear->next = l1->val < l2->val ? l1 : l2; 
        
        if(l1->val < l2->val){
            l1 = l1->next;
        }else{
            l2 = l2->next;
        }
            
        rear = rear->next;
    }
    
    if(l1){
        rear->next = l1;
        
    }
    if(l2){
        rear->next = l2;
    }
    
    
    return head.next;
    
}

struct ListNode* mergeKLists(struct ListNode** lists, int listsSize) {
    if(!listsSize){
        return NULL;
    }
    if(listsSize == 1){
        return lists[0];
    }
    
    int mid = listsSize / 2;
    Node *l1 = mergeKLists(lists, mid);
    Node *l2 = mergeKLists(lists+mid, listsSize-mid);
    
    return mergeTwoLists(l1, l2);    
}

方法二：非递归分治
非递归方法，使用原来的数组，将分治后得到的临时链表头放到数组中，每一个循环数组长度减半，
最后截止到一个元素时，即为要返回的链表头。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

typedef struct ListNode Node;

struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode head;
    head.next = NULL;
    struct ListNode *rear = &head;
    struct ListNode *node = NULL;
    
    while(l1 && l2){
        rear->next = l1->val < l2->val ? l1 : l2; 
        
        if(l1->val < l2->val){
            l1 = l1->next;
        }else{
            l2 = l2->next;
        }
            
        rear = rear->next;
    }
    
    if(l1){
        rear->next = l1;
        
    }
    if(l2){
        rear->next = l2;
    }
    
    
    return head.next;
    
}
 


struct ListNode* mergeKLists(struct ListNode** lists, int listsSize) {
    int left_size = listsSize;
    int i, k;
    int cnt;
    
    if(!listsSize){
        return NULL;
    }
    
    while(left_size>1){
        k = left_size;
        cnt = (k+1)/2;
        for(i=0; i= k ? NULL : lists[2*i+1]));
        }
        
        left_size = (left_size+1)/2;
        
    }
    return lists[0];
    
}