# Partition List Problem

## Description

LeetCode Problem 86.

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:

``````1
2
Input: head = [1,4,3,2,5,2], x = 3
Output: [1,2,2,4,3,5]
``````

Example 2:

``````1
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

## Sample C++ Code

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class Solution {
public:
ListNode* partition(ListNode* head, int x) {
return NULL;
ListNode* curr_l;
ListNode* curr_ge;

while (curr != NULL) {
if (curr->val < x) {
curr_l = curr;
} else {
curr_l->next = curr;
curr_l = curr_l->next;
}
} else {
curr_ge = curr;
} else {
curr_ge->next = curr;
curr_ge = curr_ge->next;
}
}
curr = curr->next;
}

curr_ge->next = NULL;
} else if (head_ge == NULL) {
curr_l->next = NULL;
}
else {
curr_ge->next = NULL;
}

}
};
``````