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:
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Input: head = [1,4,3,2,5,2], x = 3
Output: [1,2,2,4,3,5]
Example 2:
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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) {
if (head == NULL)
return NULL;
ListNode* head_l = NULL;
ListNode* head_ge = NULL;
ListNode* curr_l;
ListNode* curr_ge;
ListNode* curr = head;
while (curr != NULL) {
if (curr->val < x) {
if (head_l == NULL) {
head_l = curr;
curr_l = curr;
} else {
curr_l->next = curr;
curr_l = curr_l->next;
}
} else {
if (head_ge == NULL) {
head_ge = curr;
curr_ge = curr;
} else {
curr_ge->next = curr;
curr_ge = curr_ge->next;
}
}
curr = curr->next;
}
if (head_l == NULL) {
curr_ge->next = NULL;
return head_ge;
} else if (head_ge == NULL) {
curr_l->next = NULL;
return head_l;
}
else {
curr_l->next = head_ge;
curr_ge->next = NULL;
return head_l;
}
}
};