项目依赖关系分析中的数据结构

项目中常见问题

源码依赖可能导致相互直接或间接依赖形成环，对此应该如何快速检测呢？
源码依赖对于如commonLib大版本升级需要放开依赖该commonLib的其他aar源码验证，我们如何快速获取呢？

思路

针对以上问题，对于同城，同镇目前通过config维护，我们可以从该文件着手提出解决方案；项目相关的敏感信息一律以XX代替！版本信息一律为1.0.0

aar与源码切换关系由switchs集合设置：lib总计41

switchs = [
        "58AppDependenciesLib"    : dependencyMode == "aar",
        "58ClientHybridLib"       : dependencyMode == "aar",
        "58HouseLib"              : dependencyMode == "aar",
        "58HuangyeLib"            : dependencyMode == "aar",
        "58PincheLib"             : dependencyMode == "aar",
        "58PartTimeLib"           : dependencyMode == "",
        "WubaCommonsLib"          : dependencyMode == "aar",
        "WubaHybridBusinessLib"   : dependencyMode == "aar"
        ...省略...
]

若设置为源码存储位置设置在setting配置：地址为当前项目路径下

include switchs['58ClientHybridLib']?'':'58ClientHybridLib'
include switchs['58HouseLib']?'':'58HouseLib'
include switchs['58AnjukeLib'] ? '' : '58AnjukeLib'
include switchs['58CarLib']?'':'58CarLib'
 ...省略...

每个module即lib的依赖关系也在config中配置

 WubaPartTimeLib = [
            switchs['58PartTimeLib'] ? "com.wuba.XXX:" + "$WubaPartTimeLibVersion" : findProject(':58PartTimeLib'),
            rootProject.ext.WubaAppDependenciesLib,
            rootProject.ext.WubaTradelineLib,
            rootProject.ext.WubaWebBusinessLib,
            "com.wuba.XXX:$TtSdkVersion",
            'com.wuba.XXX:1.0.0',
            'com.wuba.XXX:1.0.0',
            rootProject.ext.ZCMPublish,
            rootProject.ext.WubaCommonsLib,
            rootProject.ext.WubaPermissionSDK
    ]

通过观察兼职侧依赖，有如ttsdk这种底层库，还有跟兼职一样可配置aar的module如WubaCommonsLib，其中WubaCommonsLib中也可能依赖其他可配置aar的module，如此嵌套很容易想到一种解决方式：递归调用

实施方案

首先需要获取需要获取aar与源码切换项目名称，即switchs集合数据，使用map存储
递归调用获取每个lib依赖其他lib的情况关系如下

key : WubaClientHybridLib  -- > [
com.wubaXXX:100.2.36, 
com.wubaXXXLib:100.22.41, 
com.wubanessLib:101.22.82, 
com.wubaILib:100.20.18, 
com.wubaALib:10.19.2, 
]   
key : WubaCommonsLib  -- > 
[com.wubaCommonsLib:10.26.1, 
com.wuba.SourcesLib:1.0.0, 
com.wuba.LogLib:1.0.1
]

对比以上数据，由于是递归调用当WubaClientHybridLib引用WubaCommonsLib则前者将获取到后者所有的依赖库，则我们使用何种数据格式保存他们直接的关系最为方便直接呢？
要求：方便快速查看各个lib的依赖关系，即比如给WubaCommonsLib可快速查看其依赖的底层库及它被那些lib所使用，经过对比我们选择使用图！下面简单介绍一下为何使用该方式吧！

图

定义：图是一种复杂的非线性结构，图结构中，节点之间的关系是任意的，图中任意两个数据元素之间都有可能相关，图G由两个集合V(顶点Vertex)和E(边Edge)组成，定义为G=(V，E)

还是使用上方数据举例，比如WubaClientHybridLib，WubaCommonsLib和WubaLogLib三者之间的有向图

//定义三者的顶点集合，这里使用数组也是一样儿的，数组的下标表示对应的lib名称
  String[] strArr = new String[]{
                "WubaClientHybridLib", //0 ->V0
                "WubaCommonsLib", //1->V1
                "WubaLogLib" //2 ->V2
        };

无向图：顶点的度表示以该顶点作为一个端点的边的数目；
有向图：顶点的度分为入度和出度。入度表示以该顶点为终点的入边数目，出度是以该顶点为起点的出边数目，该顶点的度等于其入度和出度之和；

image

根据以上分析，如果我们需要获取WubaClientHybridLib的依赖库只需要查找以顶点WubaClientHybridLib即顶点V0出度的边的另一个顶点。相同的如果想获取WubaLogLib被那些lib所依赖只需要获取V2入度的边的另一个顶点。

解决了如果WubaLogLib的aar被删除，可以设置只有此lib为源码依赖还是其被依赖的所有lib均设置为源码依赖，只需要读取WubaLogLib入度就可以啦！

边的表示

图的实际信息都保存在边上，它们描述了图的结构，了解数据结构的都知道，无非就是数组和链表两种形式，图也不例外：表示图的边的方法称为邻接矩阵或邻接表

邻接表

是一种顺序分配和链式分配相结合的存储结构，如这个表头结点所对应的顶点存在相邻顶点，则把相邻顶点依次存放于表头结点所指向的单向链表中；

邻接矩阵

原理就是用两个数组，一个数组保存顶点集，一个数组保存边集；

-	V1	V2
V0	1	1
V1	0	1
V2	0	0

对于每行表示当前lib依赖的其他库关系，如果为1表示存在依赖关系
对于每列表示当前lib被哪些库所依赖，如果1表示也存在依赖关系

对于config运行获取邻接矩阵关系如下，有兴趣的同学可以自己选择一个查看是否跟总结的一致；

当前key : HouseLib  -- > position :   4
当前key : PincheLib  -- > position :   12
当前key : RxDataSourcesLib  -- > position :   36
当前key : BaseUILib  -- > position :   35
当前key : VideoLib  -- > position :   31
当前key : HybridBusinessLib  -- > position :   40
当前key : FinanceLib  -- > position :   14
当前key : TribeLib  -- > position :   16
当前key : JiaoyouLib  -- > position :   25
当前key : AnjukeLib  -- > position :   20
当前key : JobCommonLib  -- > position :   8
当前key : PartTimeLib  -- > position :   13
当前key : TradelineLib  -- > position :   28
当前key : TribeABLib  -- > position :   17
当前key : TownLib  -- > position :   23
当前key : AOPLib  -- > position :   37
当前key : LogLib  -- > position :   38
当前key : WalleExtLib  -- > position :   32
当前key : IMLib  -- > position :   15
当前key : WalleLib  -- > position :   33
当前key : JobLib  -- > position :   6
当前key : HuangyeLib  -- > position :   11
当前key : RNBusinessLib  -- > position :   18
当前key : SaleLib  -- > position :   10
当前key : HouseLib  -- > position :   2
当前key : BasicBusinessLib  -- > position :   29
当前key : CarLib  -- > position :   5
当前key : LoginLib  -- > position :   21
当前key : ClientHybridABLib  -- > position :   19
当前key : TownPowerLib  -- > position :   24
当前key : LocationLib  -- > position :   22
当前key : QigsawLib  -- > position :   39
当前key : JiaoyouIMLib  -- > position :   26
当前key : JobABLib  -- > position :   7
当前key : ClientHybridLib  -- > position :   1
当前key : AppDependenciesLib  -- > position :   0
当前key : DatabaseLib  -- > position :   30
当前key : NewCarLib  -- > position :   9
当前key : WebBusinessLib  -- > position :   27
当前key : HouseAJKMixLib  -- > position :   3
当前key : CommonsLib  -- > position :   34
查看有向图表示为：int[][] map = new int[][]{            
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,0,1,0,0},
{1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,1,0,0,0,0,1,0,1,1,1,1,1,1,1,1,0,1,1,1},
{0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,1,0,0},
{0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,1,1,1,1,1,1,1,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,1,0,0},
{1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,1,1,0},
{1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,1,1,0},
{1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,1,1,1,1,1,0,1,0,0},
{1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,1,0,0},
{1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,0,1,0,0},
{1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,1,0,0},
{1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,1,1,1,1,1,1,1,0,1,0,0},
{1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,1,0,0,0,0,1,0,1,1,1,1,1,1,1,1,0,1,1,1},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,1,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,1,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,1,1,1,1,1,1,1,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,1,1,1,1,1,1,1,1,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,1,1,1,1,1,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,1,1,1,1,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,1,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,1,1,1,1,1,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,1,1,1,1,1,1,1,0,1,0,0}
};

举个例子：对于WubaClientHybridLib 下标为1，即第二列只有一个数据为1在第19行即WubaClientHybridABLib所依赖，查看config中数据果然如此

 ClientHybridABLib = [
      switchs['ClientHybridABLib'] ? "com.wuba.XXX.ClientHybridABLib:$ClientHybridABLibVersion" : findProject('ClientHybridABLib'),
      rootProject.ext.ClientHybridLib
    ]

优化点：上方邻接矩阵由于我们aar设置集合大小为41，内存占比int[41][41]大致为1681 * 4B = 6.56KB,看起来还可以接受，但随着集合数据增加，内存占比是指数级增长的，比如分析数据量为200，内存占比 200 * 200 * 4B = 156.25KB；如何减少呢？
不使用int改为byte数组，byte还是占1字节，通过观察上方数据：0表示无关，1表示存在依赖关系。那么为什么不使用bite位表示呢，对于上方41大小使用long类型占8字节 = 64bit位搓搓有余了，使用long[41]数组即可表示，内存占比64 * 8B = 0.32KB。获取相互直接关系即判断long中某一位是否为1 即Vi & (1 << (j -1)) != 0 , 判断Vi依赖关系即查询long[i]中二进制为1的位置集合，判断Vi被依赖关系即遍历long[]数组获取二进制第i 位置为1的集合；
- 注意： 若使用bite位表示将对应增加位运算量，对于数据量较小情况建议使用byte数组（仅一次读取内存操作）

int size = 41 ;
long[] nums = new long[size];
//在long数组中修改arr[i][j]存在依赖关系
nums[i] |= 1L << (size - 1 - j);
//获取long数组中i，j 处数据是否为1
nums[i] &= 1L << (size - 1 - j);
assert(nums[i] != 0)

解决

问题一

通过上方我们可以获取依赖关系的邻接矩阵的关系图，如何快速检测是否存在环呢？经常刷leetCode的同学回忆一下是否见过该问题: 207 课程表,使用两种遍历方式广度遍历BFS和深度遍历DFS

image

分析如何检测环，首先获取每个节点的入度集合degrees[n]，对于入度为0的加入栈（DFS）或者队列(BFS),这里以BFS举例；
遍历集合degrees将degrees[i]为0，即入度为0的全部节点入队列后，遍历队列取出队头元素，即断开以该节点出度的边，此时需要修改节点出度边的另一个节点的入度 - 1 后判断入度是否0，若为0加入队列，循环即可；

//储存节点的入度，统计j列中所有1的个数
int[] degrees = new int[map.length];
   for (int i = 0; i < map.length; i++) {
      for (int j = 0; j < map[0].length; j++) {
         degrees[i] += map[j][i];
      }
}
BFSSearch(map , degrees);

/**
 * BFS广度遍历：使用队列 ； DFS 深度遍历：使用栈
 * LinkedList可实现栈和队列数据：
 * 栈 ： push表示入栈，在头部添加元素 ， pop表示出栈，返回头部元素 ，peek查看栈头部元素
 * 队列： offer 尾部添加元素， poll 返回头部元素，并且从队列中删除， peek返回头部元素
 */
public void BFSSearch(int[][] map , int[] indegree){
    int count = 0; //判断有无回路（是否成环）
    LinkedList stack = new LinkedList();
    int n = indegree.length ;
    for (int i = 0; i < n; i++) {
      if (indegree[i] == 0) {
      stack.offer(i);     // --------标记点1------------
      indegree[i] = -1;
      }
    }

    while (!stack.isEmpty()) {
      Integer p = (Integer) stack.poll();  //--------标记点2------------
      System.out.print(p + " ");
      count++;
      for (int j = 0; j < n; j++) { //p列入度对应的以p出度的为p行数据
        if (map[p][j] == 1) {
            map[p][j] = 0;
            indegree[j]--;
            if (indegree[j] == 0) {
              stack.offer(j);    // --------标记点3------------
              indegree[j] = -1;
            }
          }
        }
    }
    if(count < n)
    System.out.println("当前存在回路信息!");
    else
    System.out.println("当前没有回路信息!");
}

DFS只需要简单修改几行代码即可实现

标记点1和标记点3
stack.push 入栈
标记点2
stack.pop 出栈

问题二

该问题可以提取为寻找以commonLib入度的边的集合，即统计邻接矩阵commonLib所在列值为1的集合，增加task将集合中的lib修改为源码依赖！

思考

邻接矩阵和临界表均可以表示图，我们以后用到了该如何选择，它们有什么区别呢？

邻接矩阵
- 优点： 可以快速判断两个顶点之间是否存在边（查询快）；快速添加边或者删除边；
- 缺点： 如果过多的无关联数据(稀松图)，即数组中过多0导致白白浪费空间；同时新增顶点比较麻烦，需要修改，移动数组；
临界表
- 优点： 节省空间，只存储实际存在的边。新增顶点比较简单，在入度链表中添加数据即可；
- 缺点： 关注顶点的度时，就可能需要遍历这一整个链表，有解决方式：1）增加一个反临界表存储该顶底的入度列表；2）使用十字链表表示出度，入度信息

项目依赖关系分析中的数据结构

项目中常见问题

思路

实施方案

图

边的表示

邻接表

邻接矩阵

解决

问题一

问题二

思考

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