LeetCode题解：Partition List

Partition List

Given 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.

For example,

Given

1->4->3->2->5->2

and x = 3,

return

1->2->2->4->3->5

.

思路：

遍历整个链表，把大于等于x的和小于x的数据分别放在两个不同链表中，然后再将它们连接起来。

题解：

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *partition(ListNode *head, int x) {
if (head == nullptr)
return nullptr;

ListNode* larger_head = new ListNode(0);
ListNode* larger_last = larger_head;

ListNode* less_head = new ListNode(0);
ListNode* less_last = less_head;

auto push = [](ListNode*& last, ListNode* new_less)
{
last->next = new_less;
last = last->next;
last->next = nullptr;
};

ListNode* iter = head;
while(iter != nullptr)
{
ListNode* next = iter->next;

if (iter->val < x)
push(less_last, iter);
else
push(larger_last, iter);

iter = next;
}

// merge two lists
if (less_last != less_head)
{
// less is not empty
less_last->next = larger_head->next;
delete larger_head;
less_last = less_head->next;
delete less_head;
return less_last;
}
else
{
// less is empty
delete less_head;
larger_last = larger_head;
larger_head = larger_head->next;
delete larger_last;
return larger_head;
}
}
};