medium
source: https://leetcode.com/problems/partition-list/
C/C++ Solution to LeetCode problem 86. Partition List.
Problem
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.
Examples
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
The solution is simple, we will create two list, one for numbers less than x and the other for the rest of the numbers.
- Traversing the original list, we just add each node to the correspondant list.
- At the end, we point the first list, to the head of the second list.
- Consider cases for when:
- Empty list.
- Only numbers less than
x. - Only numbers greater or equal than
x.
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/**
* 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) {
if (!head)
return head;
ListNode *less = nullptr;
ListNode *greater = nullptr;
ListNode *newHead = nullptr;
ListNode *subHead = nullptr;
ListNode *b = head;
while (b) {
if (b->val < x) {
if (!less) {
less = b;
newHead = b;
} else {
less->next = b;
less = less->next;
}
} else {
if (!greater) {
greater = b;
subHead = b;
} else {
greater->next = b;
greater = greater->next;
}
}
b = b->next;
}
if (greater)
greater->next = nullptr;
if (!less)
return subHead;
less->next = subHead;
return newHead;
}
};