725 Split Linked List in Parts

1. Question

Given a (singly) linked list with head noderoot, write a function to split the linked list intokconsecutive 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:
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Input:
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root = [1, 2, 3], k = 5
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Output: [[1],[2],[3],[],[]]
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Explanation:
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The input and each element of the output are ListNodes, not arrays.
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For example, the input root has root.val = 1, root.next.val = 2, \root.next.next.val = 3, and root.next.next.next = null.
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The first element output[0] has output[0].val = 1, output[0].next = null.
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The last element output[4] is null, but it's string representation as a ListNode is [].
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Example 2:
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Input: root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3
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Output: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]
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Explanation:
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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.
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Note:
The length ofrootwill be in the range[0, 1000].
Each value of a node in the input will be an integer in the range[0, 999].
kwill be an integer in the range[1, 50].

2. Implementation

(1) 不破坏原有的Linked List结构
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public ListNode[] splitListToParts(ListNode root, int k) {
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ListNode[] res = new ListNode[k];
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if (root == null) {
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return res;
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}
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ListNode curNode = root;
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int count = 0;
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while (curNode != null) {
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curNode = curNode.next;
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++count;
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}
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int size = count / k, rem = count % k;
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curNode = root;
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for (int i = 0; i < k; i++) {
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ListNode dummy = new ListNode(0);
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ListNode temp = dummy;
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for (int j = 0; j < size + (i < rem ? 1 : 0); j++) {
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temp.next = new ListNode(curNode.val);
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temp = temp.next;
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curNode = curNode.next;
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}
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res[i] = dummy.next;
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}
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return res;
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}
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}
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(2) 破坏原有的Linked List结构
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public ListNode[] splitListToParts(ListNode root, int k) {
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ListNode[] res = new ListNode[k];
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if (root == null) {
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return res;
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}
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ListNode curNode = root;
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int count = 0;
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while (curNode != null) {
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curNode = curNode.next;
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++count;
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}
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int size = count / k, rem = count % k;
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curNode = root;
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for (int i = 0; i < k; i++) {
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ListNode head = curNode;
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// 减一是因为我们到后面要切开Linked List的部分结构存入res里
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for (int j = 0; j < size + (i < rem ? 1 : 0) - 1; j++) {
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if (curNode != null) {
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curNode = curNode.next;
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}
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}
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// 切开Linked List
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if (curNode != null) {
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ListNode preNode = curNode;
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curNode = curNode.next;
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preNode.next = null;
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}
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res[i] = head;
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}
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return res;
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}
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}
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3. Time & Space Complexity

不破坏原有的Linked List结构: 时间复杂度O(n), 空间复杂度O(Max(n, k))
破坏原有的Linked List结构: 时间复杂度O(n), 空间复杂度O(k)