# 817. Linked List Components 链表组件

@TOC

## # 题目描述

We are given `head`, the head node of a linked list containing unique integer values.

We are also given the list `G`, a subset of the values in the linked list.

Return the number of connected components in `G`, where two values are connected if they appear consecutively in the linked list.

Example 1:

``````Input:
G = [0, 1, 3]
Output: 2
Explanation:
0 and 1 are connected, so [0, 1] and [3] are the two connected components.
``````

Example 2:

``````Input:
G = [0, 3, 1, 4]
Output: 2
Explanation:
0 and 1 are connected, 3 and 4 are connected, so [0, 1] and [3, 4] are the two connected components.
``````

Note:

1. If N is the length of the linked list given by head, 1 <= N <= 10000.
2. The value of each node in the linked list will be in the range [0, N - 1].
3. 1 <= G.length <= 10000.
4. G is a subset of all values in the linked list.

## # 解题方法

``````# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
"""
:type G: List[int]
:rtype: int
"""
groups = 0
subset = set(G)
groups += 1
return groups
``````

C++版本的代码如下：

``````/**
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
int numComponents(ListNode* head, vector<int>& G) {
set<int> s(G.begin(), G.end());
int res = 0;
bool isCon = false;
while (p) {
if (s.count(p->val)) {
if (!isCon) {
res ++;
isCon = true;
}
} else {
isCon = false;
}
p = p->next;
}
return res;
}
};
``````

## # 日期

2018 年 5 月 28 日 —— 太阳真的像日光灯～ 2018 年 12 月 14 日 —— 12月过半，2019就要开始