46. 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.

分析： 保持两个表头, 一个存大于等于 x 的结点，一个存小于 x 的结点。

注意：记住两个表尾，中间不能将其 next 指针设置为空。最后将存小值的表尾链接在另一个前面，两一个表尾 next 置为 NULL.

/**
 * 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 *head1, *head2, *tail1, *tail2;
        head1 = head2 = tail1 = tail2 = NULL;
        while(head) {
            if(head->val >= x) {
                if(head2 == NULL) head2 = tail2 = head;
                else { tail2->next = head; tail2 = tail2->next; }
            } else {
                if(head1 == NULL) head1 = tail1 = head;
                else { tail1->next = head; tail1 = tail1->next; }
            }
            head = head->next;
        }
        if(head2) tail2->next = NULL;
        if(head1) tail1->next = head2;
        else head1 = head2;
        return head1;
    }
};