[Leetcode] Partition List

最新推荐文章于 2019-07-16 09:33:43 发布

原创最新推荐文章于 2019-07-16 09:33:43 发布 · 117 阅读

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本文详细阐述了如何通过链表节点值与给定值x进行比较，实现链表的分区操作，确保所有小于x的节点位于前面，大于等于x的节点位于后面，同时保持原始相对顺序不变。

Partition List

AC Rate: 428/1694

My Submissions

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.

/**
 * 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 lessHead(99), largeHead(99);
        ListNode* less = &lessHead, *large = &largeHead, *cur = head;
        while (cur != NULL) {
            if (cur->val < x) {
                less->next = cur;
                less = cur;
            } else {
                large->next = cur;
                large = cur;
            }
            cur = cur->next;
        }
        large->next = NULL;
        less->next = largeHead.next;
        return lessHead.next;
    }
};