leetcode解题: Partition List (86)

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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.

解法1：Dummy Node, O(N) One pass

这题比array的partition容易一些，主要用两个dummy node记录两个list，一个小于x，一个大于等于x。用一个head指针维护现在遍历到的node。最后将两个dummy连在一起就可以了。
C++

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/**
 * 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 = new ListNode(0);
        ListNode* lefttail = left;
        ListNode* right = new ListNode(0);
        ListNode* righttail = right;
        while (head != NULL) {
            ListNode* next = head->next;
            head->next = NULL;
            if (head->val < x) {
                lefttail->next = head;
                lefttail = lefttail->next;
            } else {
                righttail->next = head;
                righttail = righttail->next;
            }
            head = next;
        }
        lefttail->next = right->next;
        ListNode* res = left->next;
        delete left;
        delete right;
        return res;
    }
};

Java

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