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Copy path725.SplitListToParts.cs
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725.SplitListToParts.cs
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// 725. Split Linked List in Parts
// Given the head of a singly linked list and an integer k, 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 one. This may lead to some parts being null.
// The parts should be in the order of occurrence in the input list, and parts occurring earlier should always have a size greater than or equal to parts occurring later.
// Return an array of the k parts.
// Example 1:
// Input: head = [1,2,3], k = 5
// Output: [[1],[2],[3],[],[]]
// Explanation:
// The first element output[0] has output[0].val = 1, output[0].next = null.
// The last element output[4] is null, but its string representation as a ListNode is [].
// Example 2:
// Input: head = [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.
/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int val=0, ListNode next=null) {
* this.val = val;
* this.next = next;
* }
* }
*/
public class Solution {
public ListNode[] SplitListToParts(ListNode head, int k) {
int len = 0;
ListNode curr = head;
while(curr!=null){
len++;
curr = curr.next;
}
int arrLen = len/k;
int mod = len % k;
ListNode[] result = new ListNode[k];
for(int i = 0; i < k; i++){
result[i] = head;
ListNode prev = head;
int currLen = i < mod ? arrLen + 1 : arrLen;
while(head != null && currLen > 0){
prev = head;
head = head.next;
currLen--;
}
if(prev!=null)
prev.next = null;
}
return result;
}
}