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partition_list.cpp
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52 lines (42 loc) · 1.31 KB
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/*
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.
*/
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *partition(ListNode *head, int x) {
ListNode dummy1(0);
ListNode dummy2(0);
ListNode *tail1 = &dummy1;
ListNode *tail2 = &dummy2;
while (head) {
ListNode **tail;
if (head -> val < x)
tail = &tail1;
else
tail = &tail2;
(*tail) -> next = head;
head = head -> next;
*tail = (*tail) -> next;
(*tail) -> next = NULL;
}
tail1 -> next = dummy2.next;
return dummy1.next;
}
};
/*
The in-place partition operation for linked lists is much simpler than it for arrays.
Can you think of other operations that are simpler/faster on linked lists than they on arrays?
rotate operation, merge sort (because sorted_merge on linked lists doesn't require additional buffer)
*/