Partition List
Problem Statement :
Given the head of 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. Example 1: Input: head = [1,4,3,2,5,2], x = 3 Output: [1,2,2,4,3,5] Example 2: Input: head = [2,1], x = 2 Output: [1,2] Constraints: The number of nodes in the list is in the range [0, 200]. -100 <= Node.val <= 100 -200 <= x <= 200
Solution :
Solution in C :
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* partition(struct ListNode* head, int x) {
struct ListNode* dummy = (struct ListNode*)malloc(sizeof(struct ListNode));
dummy->val = 0;
dummy->next = NULL;
struct ListNode* prev = dummy;
struct ListNode* greater = (struct ListNode*)malloc(sizeof(struct ListNode));
greater->val = 0;
greater->next = NULL;
struct ListNode* last = greater;
while(head != NULL) {
if(head->val < x) {
prev->next = head;
prev = prev->next;
} else {
last->next = head;
last = last->next;
}
head = head->next;
}
last->next = NULL;
prev->next = greater->next;
struct ListNode *newHead = dummy->next;
free(dummy);
free(greater);
return newHead;
}
Solution in C++ :
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* partition(ListNode* head, int x) {
ListNode* dummy = new ListNode(0);
ListNode* prev = dummy;
ListNode* greater = new ListNode(0);
ListNode* last = greater;
while(head != NULL){
if(head->val < x){
prev->next = head;
prev = prev->next;
}else{
last->next = head;
last = last->next;
}
head = head->next;
}
last->next = NULL;
prev->next = greater->next;
return dummy->next;
}
};
Solution in Java :
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode partition(ListNode head, int x) {
ListNode dummy = new ListNode(0);
ListNode front = dummy;
ListNode dummy2 = new ListNode(0);
ListNode back = dummy2;
while(head != null){
if(head.val < x){
front.next = head;
front = front.next;
}else{
back.next = head;
back = back.next;
}
head = head.next;
}
back.next = null;
front.next = dummy2.next;
return dummy.next;
}
}
Solution in Python :
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution(object):
def partition(self, head, x):
"""
:type head: ListNode
:type x: int
:rtype: ListNode
"""
dummy = ListNode(0)
front = dummy
dummy2 = ListNode(0)
back = dummy2
while head != None :
if head.val < x :
front.next = head
front = front.next
else :
back.next = head
back = back.next
head = head.next
back.next = None
front.next = dummy2.next
return dummy.next
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