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PartitionList.h
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/*
Author: naiyong, aonaiyong@gmail.com
Date: Sep 22, 2014
Problem: Partition List
Difficulty: 3
Source: https://github.com/aonaiyong/LeetCode
Notes:
Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.
You should preserve the original relative order of the nodes in each of the two partitions.
For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.
Solution: 1. First decouple nodes into two lists, and then splice the lists.
2. Partition in place.
*/
#ifndef PARTITIONLIST_H_
#define PARTITIONLIST_H_
#include "ListNode.h"
class Solution {
public:
ListNode *partition(ListNode *head, int x) {
return partitionDecouple(head, x);
}
// succinct but maybe less efficient
ListNode *partitionDecouple(ListNode *head, int x) {
ListNode dummy1(0), *tail1 = &dummy1;
ListNode dummy2(0), *tail2 = &dummy2;
// decouple nodes < x and nodes >= x, respectively
while (head) {
if (head->val < x)
tail1 = tail1->next = head;
else
tail2 = tail2->next = head;
head = head->next;
}
// splice the two lists
tail1->next = dummy2.next;
tail2->next = nullptr; // important
return dummy1.next;
}
// verbose
ListNode *partitionInPlace(ListNode *head, int x) {
ListNode dummy(0), *less = &dummy, *greater = &dummy;
dummy.next = head;
while (greater->next) {
ListNode *curr = greater->next;
if (x <= curr->val)
greater = curr;
else {
if (less == greater) // "greater part" is empty
less = greater = curr;
else {
greater->next = curr->next;
curr->next = less->next;
less = less->next = curr;
}
}
}
return dummy.next;
}
};
#endif /* PARTITIONLIST_H_ */