reversing linked list

项目场景：

提示：这里简述

Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K=3, then you must output 3→2→1→6→5→4; if K=4, you must output 4→3→2→1→5→6.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (≤105) which is the total number of nodes, and a positive K (≤N) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.

Then N lines follow, each describes a node in the format:

Address Data Next

where Address is the position of the node, Data is an integer, and Next is the position of the next node.

Output Specification:

For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218

结尾无空行

Sample Output:

00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1

结尾无空行

---

实现思路：

#include <iostream>
#include <iomanip>
#include <vector>
using namespace std;

const int Maxsize = 100005;

struct node {
	int data = 0;
	int next = 0;
};
int main() {
	node List[Maxsize];
	int head, N, K, t, flag;//flag记录count数组中有效节点的数量
	cin >> head >> N >> K;//输入头节点地址，节点数，反转数
	vector<int> count(N);

	for (int i = 0;i < N;i++) {//输入每个节点的数据及下一个节点的地址
		int  address;
		cin >> address;
		cin >> List[address].data;
		cin >> List[address].next;
	}
	flag = 0;   // 累计有效结点数 
	while (head != -1) {   // 当尾结点为 -1 时结束 
		count[flag++] = head;   // 记录所有Address
		head = List[head].next;  // 找下一个结点 
	}


	for (int i = 0;i < flag - flag % K&&N>1;i = +K) {
		for (int j = 0;j < K / 2;j++) {
			t = count[i + K - j - 1];
			count[i + K - 1 - j] = count[i + j];
			count[i + j] = t;
		}
	}

	for (int i = 0;i < flag - 1;i++) {
		cout << setw(5) << setfill('0') << count[i] << " " << List[count[i]].data << " ";
		cout << setw(5) << setfill('0') << List[count[i]].next << endl;
	}
	cout << setw(5) << setfill('0') << count[flag - 1] << " " << List[count[flag - 1]].data << " ";
	cout << List[count[flag - 1]].next;
	return 0;
}

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原因分析：

提示：这里填写问题的分析：
例如：Handler 发送消息有两种方式，分别是 Handler.obtainMessage()和 Handler.sendMessage()，其中 obtainMessage 方式当数据量过大时，由于 MessageQuene 大小也有限，所以当 message 处理不及时时，会造成先传的数据被覆盖，进而导致数据丢失。

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解决方案：

提示：这里填写该问题的具体解决方案：
例如：新建一个 Message 对象，并将读取到的数据存入 Message，然后 mHandler.obtainMessage(READ_DATA, bytes, -1, buffer).sendToTarget();换成 mHandler.sendMessage()。