725. Split Linked List in Parts

题目意思很简单，给你一串链表，将其分成k个链表。前面链表的长度要大于后面链表的长度。比如说长度为10分成3部分就是[4,3,3]

Example:

Input:
root = [1, 2, 3], k = 5

Output: [[1],[2],[3],[],[]]

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].

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    vector<ListNode*> splitListToParts(ListNode* root, int k) {
        vector<ListNode*> res;
        ListNode* tmp = root;
        int cnt = 0;
        while(tmp!=NULL){
            cnt++;
            tmp = tmp->next;
        } 
        int per_cnt[55] = {0};
        for(int i = 0; i < k; i++){
            per_cnt[i] = cnt/k;
        }
        //余数分给前面的链表
        for(int i=0; i < cnt % k; i++){
            per_cnt[i] += 1;
            
        }
       
        tmp = root;
        for(int i = 0; i < k; i++){
            ListNode* lastPtr = NULL;
            ListNode* p = NULL;
            for(int j=0;j<per_cnt[i];j++){
                if(tmp == NULL)
                    break;
                if(j == 0){//链表的头存在p指针中,lastPtr指针指向链表现在所在的位置
                    lastPtr = new ListNode(tmp->val);
                    p = lastPtr;
                    tmp = tmp -> next;
                    continue;
                }
                ListNode* tmpNode = new ListNode(tmp->val);
                lastPtr -> next = tmpNode;//往当前指针插入下一个节点
                lastPtr = tmpNode;//当前指针跑到下一个节点去
                tmp = tmp -> next;
            }
            res.push_back(p);
        }
        return res;
    }
};