LeetCode.725 Split Linked List in Parts (将链表分成k段)
1.题目
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 ]
2.样例
- 1.样例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 [].
- 2.样例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.
-
3.notes
- 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].
3.代码及思路
- 思路：对链表进行均分为k段，将多余的余数分别从链表头分发至每段链表即可。
- 代码：
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public static ListNode[] splitListToParts(ListNode root, int k) {
// 思路：均分即可，多出来的余数平均分配给前面每一个，ps：需要考虑空root的情况
ListNode tmp = root;
int len = 0;
while (tmp != null) {
len++;
tmp = tmp.next;
}
// 长度相同均为1
List<Integer> lenList = new ArrayList<>();
if (len <= k) {
for (int i = 1; i <= k; i++) {
lenList.add(1);
}
} else {
int ave = len / k;
if (len % k == 0) {
// 均分
for (int i = 1; i <= k; i++) {
lenList.add(ave);
}
} else {
// 存在余数
int remain = len % k;
for (int i = 1; i <= k; i++) {
if (i <= remain) {
lenList.add(ave + 1);
} else {
lenList.add(ave);
}
}
}
}
// 切分数组
return splitArr(root, lenList);
}
public static ListNode[] splitArr(ListNode root, List<Integer> lenList) {
ListNode[] res = new ListNode[lenList.size()];
for (int i = 0; i < lenList.size(); i++) {
int len = lenList.get(i);
ListNode headNode = new ListNode(-1);
ListNode curNode = headNode;
while (len-- > 0) {
if (root != null) {
curNode.next = new ListNode(root.val);
curNode = curNode.next;
root = root.next == null ? null : root.next;
}
}
res[i] = headNode.next;
}
return res;
}
}