Leetcode: Partition List

思路:

1. 空间复杂度为 o(n) 解法. 创建两个链表, 分别记录大于 x 和小于 x 的节点, 最后合并

2. o(1) 的空间复杂度解法. 四个指针, 分别指向小于 x 部分链表的头, 尾, 指向大于 x 部分链表的头, 尾

总结:

1. 使用 dummyNode 减少判断

代码:

class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        ListNode *first, *second, *third, *fourth;
		if(head == NULL)
			return head;

		ListNode *cursor = head;
		ListNode *dummyNode = new ListNode(0);
		dummyNode->next = head;
		first = second =  dummyNode;
		while(cursor && cursor->val < x) { 
			second = cursor;
			cursor = cursor->next;
		}
		if(cursor && cursor->val >= x) {
			third = cursor;
			fourth = cursor;
		}
		while(cursor) {
			if(cursor->val < x) {
				second->next = cursor;
				fourth->next = cursor->next;
				cursor->next = third;
				second = cursor;
			}else{
				fourth = cursor;
			}
			cursor = cursor->next;
		}
		return dummyNode->next;
    }
};

　　

Update

较为简洁的代码

class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        if(head == NULL)
            return head;

        ListNode *sm = new ListNode(0), *sm_tail = sm;
        ListNode *bg = new ListNode(0), *bg_tail = bg;
        ListNode *cur = head;

        while(cur) {
            if(cur->val < x) {
                sm_tail->next = cur;
                cur     = cur->next;
                sm_tail = sm_tail->next;
                sm_tail->next = NULL;
            }else{
                bg_tail->next = cur;
                bg_tail = bg_tail->next;
                cur = cur->next;
                bg_tail->next = NULL;
            }
        }
        sm_tail->next = bg->next;
        return sm->next;
    }
};