Partition List

Source

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->null and x = 3,
return 1->2->2->4->3->5->null.

题解

依据题意，是要根据值x对链表进行分割操作，具体是指将所有小于x的节点放到不小于x的节点之前，咋一看和快速排序的分割有些类似，但是这个题的不同之处在于只要求将小于x的节点放到前面，而并不要求对元素进行排序。

这种分割的题使用两路指针即可轻松解决。左边指针指向小于x的节点，右边指针指向不小于x的节点。由于左右头节点不确定，我们可以使用两个dummy节点。

C++

/**
 * 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) {
        if (head == NULL) return NULL;

        ListNode *leftDummy = new ListNode(0);
        ListNode *left = leftDummy;
        ListNode *rightDummy = new ListNode(0);
        ListNode *right = rightDummy;
        ListNode *node = head;
        while (node != NULL) {
            if (node->val < x) {
                left->next = node;
                left = left->next;
            } else {
                right->next = node;
                right = right->next;
            }
            node = node->next;
        }
        // post-processing
        right->next = NULL;
        left->next = rightDummy->next;

        return leftDummy->next;
    }
};

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
        if (head == null) return null;

        ListNode leftDummy = new ListNode(0);
        ListNode left = leftDummy;
        ListNode rightDummy = new ListNode(0);
        ListNode right = rightDummy;
        ListNode node = head;
        while (node != null) {
            if (node.val < x) {
                left.next = node;
                left = left.next;
            } else {
                right.next = node;
                right = right.next;
            }
            node = node.next;
        }
        // post-processing
        right.next = null;
        left.next = rightDummy.next;

        return leftDummy.next;
    }
}

源码分析

1. 异常处理
2. 引入左右两个dummy节点及left和right左右尾指针
3. 遍历原链表
4. 处理右链表，置right->next为空，将右链表的头部链接到左链表尾指针的next，返回左链表的头部

复杂度分析

遍历链表一次，时间复杂度近似为 O(n), 使用了两个 dummy 节点及中间变量，空间复杂度近似为 O(1).