下载量已超越千万的爆红游戏2048（同名小3传奇、1024）源代码解密和下载（第二篇）

简介上一篇我们主要分析了1024原版的基本思路和简单算法，这一篇我们将继续分析如何实现各种版本的聚合；运行demo需要配置好CocosEditor，暂不支持其他工具。demo基于javascript语言，cocos2d-x游戏引擎，CocosEditor手游开发工具完成的，可移植运行android，ios，html5网页等。apk演示效果360应用市场（2048 聚合版）：http://zhushou.360.cn/detail/index/soft_id/1634607?recrefer=SE_D_2048%20%E8%81%9A%E5%90%88
豌豆荚应用市场 ：http://www.wandoujia.com/apps/com.touchsnow.game.sudoku
CocosEditor版源代码下载：
cocos2d-js源代码请到集中营下载：http://blog.makeapp.co/?p=523

效果图：






代码分析
1 建立JS数组，有三个数组 背景颜色-等级分数-版本选择 建好之后可以在程序里面直接调用；
版本数组里面每一个版本名字、类型、字体大小和等级数组

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COLOR = [cc.c3b(255, 255, 255),

cc.c3b(238, 246, 246), cc.c3b(172, 141, 173), cc.c3b(255, 237, 196),

cc.c3b(242, 197, 170), cc.c3b(191, 164, 157), cc.c3b(125, 125, 106),

cc.c3b(247, 240, 145), cc.c3b(221, 204, 163), cc.c3b(251, 96, 191), cc.c3b(254, 128, 128),

cc.c3b(211, 84, 113), cc.c3b(63, 130, 211),

//other

cc.c3b(100, 85, 39), cc.c3b(75, 34, 255), cc.c3b(58, 68, 104),

cc.c3b(7, 0, 234), cc.c3b(153, 45, 85), cc.c3b(15, 254, 36), cc.c3b(78, 2, 34),

cc.c3b(255, 125, 64), cc.c3b(237, 145, 33)

];

SCORES = [0, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048];

VERSIONS = [

{

num: 1,

name: "2048原版",

type: "number",

labelFontSize: 50,

array: [0, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048]

},

{

num: 2,

name: "历史版",

type: "string",

labelFontSize: 50,

array: [" ", "商", "周", "秦", "汉", "唐", "宋", "元", "明", "清", "民国", "天朝"]

},

{

num: 3,

name: "后宫版",

type: "string",

labelFontSize: 40,

array: [" ", "宫女", "答应", "常在", "贵人", "嫔", "妃", "贵妃", "皇贵妃", "皇后", "皇太后", "太皇太后"]

},

{

num: 4,

name: "军旅版",

type: "string",

labelFontSize: 40,

array: [" ", "小兵", "班长", "排长", "连长", "营长", "团长", "旅长", "师长", "军长", "司令", "军委主席"]

},

{

num: 5,

name: "学历版",

type: "string",

labelFontSize: 40,

array: [" ", "幼儿", "小学", "中学", "高中", "专科", "本科", "研究", "硕士", "博士", "博士后", "院士"]

},

{

num: 6,

name: "金庸版",

type: "string",

labelFontSize: 40,

array: [" ", "令狐冲", "杨过", "郭靖", "虚竹", "风清扬", "张三丰", "东方不败", "逍遥老祖", "独孤求败", "扫地神僧", "达摩老祖"]

},

{

num: 7,

name: "生肖版",

type: "string",

labelFontSize: 40,

array: [" ", "鼠", "牛", "虎", "兔", "龙", "蛇", "马", "羊", "猴", "鸡", "狗", "猪", "龙"]

},

{

num: 8,

name: "星座版",

type: "string",

labelFontSize: 40,

array: [" ", "白羊座", "金牛座", "双子座", "巨蟹座", "狮子座", "处女座", "天秤座", "天蝎座", "射手座", "魔羯座", "水瓶座", "双鱼座"]

},

{

num: 9,

name: "甲乙丙版",

type: "string",

labelFontSize: 40,

array: [" ", "甲", "乙", "丙", "丁", "戊", "己", "庚", "辛", "壬", "癸"]

},

{

num: 10,

name: "豪车版",

type: "string",

labelFontSize: 40,

array: [" ", "大众", "奔驰", "悍马", "法拉利", "奥迪", "宝马", "兰博基尼", "凯迪拉克", "世爵"]

}

];

2 然后找到建好的版本选择LevelLayer.js，初始化4*4版本选择表格和点击单个表格跳转到不同的版本里面
#把plist加入缓存addSpriteFrames（）；
#双层循环this.tables,每一个单元格放入一个新建的精灵cell=cc.MySprite.create(this.rootNode, "6.png", cc.p(px, py), 1)
#根据i，j确定cell的编号num=4*i+j；
#有了编号num可以从VERSIONS数组里面获取版本名字VERSIONS[num].name
#建立文字标签cellLabel = cc.MySprite.createLabel(cell, VERSIONS[num].name);

#表格初始化之后需要添加触摸事件；又一次双层循环this.tables
#检测精灵触摸cc.rectContainsPoint(sprite.getBoundingBox(), loc）
#找到编号num，传值到全局变量indexVersions=num；主场景就根据这个标示来确定版本；
#跳转到主场景cc.BuilderReader.runScene("", "MainLayer");

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LevelLayer.prototype.onEnter = function () {

cc.SpriteFrameCache.getInstance().addSpriteFrames("res/main.plist");

this.initX = 9;

this.initY = 850;

this.cellSize = 162;

this.cellSpace = 18;

this.tables = new Array(4);

for (var j = 0; j < 4; j++) {

var sprites = new Array(4);

for (var i = 0; i < 4; i++) {

var px = this.initX + this.cellSize / 2 + i * (this.cellSize + this.cellSpace);

var py = this.initY + this.cellSize / 2 - j * (this.cellSize + this.cellSpace);

var cell = cc.MySprite.create(this.rootNode, "6.png", cc.p(px, py), 1);

var num = 4 * j + i;

if (num < VERSIONS.length) {

var cellLabel = cc.MySprite.createLabel(cell, VERSIONS[num].name);

cellLabel.setFontSize(30);

}

sprites[i] = cell;

}

this.tables[j] = sprites;

}

};

LevelLayer.prototype.onTouchesBegan = function (touches, event) {

cc.log("onTouchesBegan");

var loc = touches[0].getLocation();

for (var j = 0; j < 4; j++) {

for (var i = 0; i < 4; i++) {

var sprite = this.tables[j][i];

if (cc.rectContainsPoint(sprite.getBoundingBox(), loc)) {

var num = 4 * j + i;

cc.log("num==" + num);

if (num < VERSIONS.length) {

indexVersions = num;

cc.BuilderReader.runScene("", "MainLayer");

}

}

}

}

};

3 进入主场景，根据第二步得到的indexVersions，从数组VERSIONS里面获取版本实体this.indexVersion = VERSIONS[this.versionNum];
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MainLayer.prototype.onEnter = function () {

//version

this.versionNum = indexVersions;

this.indexVersion = VERSIONS[this.versionNum];

this.title.setString(this.indexVersion.name + "目标：" + this.indexVersion.array[this.indexVersion.array.length - 1] + "");}

4 刷新按钮事件和返回按钮；刷新按钮比较简单，关键是返回按钮，它是一个文字标签，cocos里面没有关联点击事件；这里笔者通过触摸和创建矩形的方法来解决了这个问题；

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MainLayer.prototype.onRefreshClicked = function () {

cc.BuilderReader.runScene("", "MainLayer");

};

MainLayer.prototype.onTouchesBegan = function (touches, event) {

this.pBegan = touches[0].getLocation();

//back

var backRect = cc.rectCreate(this.back.getPosition(), [50, 30]);

if (cc.rectContainsPoint(backRect, this.pBegan)) {

this.back.runAction(cc.Sequence.create(cc.ScaleTo.create(0.2, 1.1),

cc.CallFunc.create(function () {

cc.AudioEngine.getInstance().stopAllEffects();

cc.BuilderReader.runScene("", "LevelLayer");

})

));

}

};

两篇博文完结；