标签：arrays linked 资料 html 参考 fun jimmy while order
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 [ [1], [2], [3], [4], null ]Example 1:
Input: root = [1, 2, 3], k = 5 Output: [[1],[2],[3],[],[]] 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[0] has output[0].val = 1, output[0].next = null. The last element output[4] 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.
Java Solution:
Runtime beats 59.09%
完成日期：12/07/2017
关键词：singly-linked list
关键点：计算出base 和 leftover
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { val = x; } 7 * } 8 */ 9 class Solution 10 { 11 public ListNode[] splitListToParts(ListNode root, int k) 12 { 13 ListNode[] list = new ListNode[k]; 14 int len = 0; 15 ListNode cursor = root; 16 int base = 0; 17 int leftover = 0; 18 19 // count the length 20 while(cursor != null) 21 { 22 cursor = cursor.next; 23 len++; 24 } 25 26 // calculate the base and leftover 27 base = len / k; 28 leftover = len % k; 29 cursor = root; 30 31 // iterate each group 32 for(int i=0; i<k; i++) 33 { 34 list[i] = cursor; // save this group head 35 ListNode tail = null; 36 int groupSize = base; 37 38 // set up correct group size 39 if(leftover > 0) 40 { 41 groupSize++; 42 leftover--; 43 } 44 45 // iterate this group nodes 46 for(int j=0; j<groupSize; j++) 47 { 48 if(j == groupSize - 1) // approach to the end of this group 49 tail = cursor; 50 51 cursor = cursor.next; 52 } 53 54 if(groupSize > 0) // link this group tail to null 55 tail.next = null; 56 } 57 58 59 return list; 60 } 61 }
参考资料：N/A
LeetCode 题目列表 - LeetCode Questions List
题目来源：https://leetcode.com/
LeetCode 725. Split Linked List in Parts (分裂链表)
标签：arrays linked 资料 html 参考 fun jimmy while order
原文地址：http://www.cnblogs.com/jimmycheng/p/8004638.html
