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.
解题思路:
遍历链表,小于x的节点归为left链表,大于等于x的节点归为right链表,合并两个链表。
/**
* 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) {
ListNode left_dummy(-1);
ListNode right_dummy(-1);
auto left_cur = &left_dummy;
auto right_cur = &right_dummy;
for (ListNode *cur = head; cur; cur = cur->next) {
if (cur->val < x) {
left_cur->next = cur;
left_cur = cur;
}
else {
right_cur->next = cur;
right_cur = cur;
}
}
left_cur->next = right_dummy.next;
right_cur->next = NULL;
return left_dummy.next;
}
};