[Leetcode] 70. Partition List

题目

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.

解题之法

/**
 * 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) return head;
        ListNode *dummy = new ListNode(-1);
        ListNode *newDummy = new ListNode(-1);
        dummy->next = head;
        ListNode *cur = dummy, *p = newDummy;
        while (cur->next) {
            if (cur->next->val < x) {
                p->next = cur->next;
                p = p->next;
                cur->next = cur->next->next;
                p->next = NULL;
            } else {
                cur = cur->next;
            }
        }
        p->next = dummy->next;
        return newDummy->next;
    }
};

分析

这道题要求我们划分链表，把所有小于给定值的节点都移到前面，大于该值的节点顺序不变，相当于一个局部排序的问题。
可以将所有小于给定值的节点取出组成一个新的链表，此时原链表中剩余的节点的值都大于或等于给定值，只要将原链表直接接在新链表后即可。