# [leetcode-链表]725. Split Linked List in Parts

xiaoxiao2021-03-01  12

Given a (singly) linked list with head node root, write a function to split the linked list into k consecutive linked list "parts".

The length of each part should be as equal as possible: no two parts should have a size differing by more than 1. This may lead to some parts being null.

The parts should be in order of occurrence in the input list, and parts occurring earlier should always have a size greater than or equal parts occurring later.

Return a List of ListNode's representing the linked list parts that are formed.

Examples 1->2->3->4, k = 5 // 5 equal parts [ , , , , null ]

Example 1:

Input: root = [1, 2, 3], k = 5 Output: [,,,[],[]] Explanation: The input and each element of the output are ListNodes, not arrays. For example, the input root has root.val = 1, root.next.val = 2, \root.next.next.val = 3, and root.next.next.next = null. The first element output has output.val = 1, output.next = null. The last element output is null, but it's string representation as a ListNode is [].

Example 2:

Input: root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3 Output: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]] Explanation: The input has been split into consecutive parts with size difference at most 1, and earlier parts are a larger size than the later parts.

Note:

The length of root will be in the range [0, 1000].Each value of a node in the input will be an integer in the range [0, 999].k will be an integer in the range [1, 50].

//================================================================================

/**  * Definition for singly-linked list.  * struct ListNode {  *     int val;  *     struct ListNode *next;  * };  */ /**  * Return an array of size *returnSize.  * Note: The returned array must be malloced, assume caller calls free().  */ struct ListNode** splitListToParts(struct ListNode* root, int k, int* returnSize) {         struct ListNode *pcur = root;          //计算链表长度      int len = 0;     while(pcur)     {          ++len;         pcur = pcur->next;     }             //计算每个子链表的长度（长度不够的，分配1个元素NULL)      int parts_num = (len < k) ? 1 : (len/k);          //计算前多少个子链表会多分配1个元素      int extra = (len < k) ? (0) : (len % k);

//分配k个子链表的数组(calloc自动清零）      struct ListNode** lists = calloc(k , sizeof(struct ListNode*));     //struct ListNode** lists = (struct ListNode **)malloc(k * sizeof(struct ListNode*));     //memset(lists, 0, k*sizeof(struct ListNode*));     *returnSize = k;          //从原本的链表头开始遍历      struct ListNode *prev=NULL;     pcur = root;

int i, j;     for (i = 0; i < k; i++, --extra )//k个子链表，前extra个子链表元素个数多1个      {         //把每个子链表头放进链表数组里          lists[i] = pcur;                 //给每个子链表分配元素          for (j = 0 ; j < parts_num + (extra > 0) && pcur ; j++)          {             prev = pcur;                   pcur = pcur->next;         }                  //每个子链表彼此断开          if (prev)              prev->next = NULL; //breaking list became a independent list.     }

return lists; }