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.
/**
* 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);
ListNode* left_cur = &left_dummy;
ListNode* 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;
}
};版权声明：本文为博主原创文章，未经博主允许不得转载。