86. Partition List
1. Description
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.
2. Example
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]
3. Constraints
- The number of nodes in the list is in the range [0, 200].
- -100 <= Node.val <= 100
- -200 <= x <= 200
4. Solutions
My Accepted Solution
n is the number of nodes in i_head
Time complexity: O(n)
Space complexity: O(1)
class Solution {
public:
// ListNode* partition(ListNode* head, int x)
ListNode* partition(ListNode* i_head, const int partition) {
ListNode* littleHead = new ListNode();
ListNode* BigHead = new ListNode();
auto littleIter = littleHead, bigIter = BigHead;
for (auto iter = i_head; iter; iter = iter->next) {
if (iter->val < partition) {
littleIter->next = iter;
littleIter = littleIter->next;
} else {
bigIter->next = iter;
bigIter = bigIter->next;
}
}
littleIter->next = BigHead->next;
bigIter->next = nullptr;
return littleHead->next;
}
};