LeetCode Partition List

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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.

思路：不难的链表问题，处理一下就行了，两个链表辅助。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
        if (head == null) return null;
        
        ListNode lessNode = new ListNode(0);
        ListNode greaterNode = new ListNode(0);
        ListNode cur = head, less = lessNode, greater = greaterNode;
        while (cur != null) {
        	ListNode next = cur.next;
        	if (cur.val < x) {
        		less.next = cur;
        		less = less.next;
        		less.next = null;
        	} else {
        		greater.next = cur;
        		greater = greater.next;
        		greater.next = null;
        	}
        	cur = next;
        }
        
        less.next = greaterNode.next;
        return lessNode.next;
    }
}

LeetCode Partition List

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原文地址：http://blog.csdn.net/u011345136/article/details/45009833