寒風的Cocos之旅之Lua篇---2048移动块的机制

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用Lua又写了遍2048.这回发现比之前写的更快了。在上下左右移动块的上边，例如像左移动：那么需要遍历的则

是x轴方向。并且是跟移动方向相反的向右进行遍历。此时定义一个对象，记录向右遍历的第i个的值，记录后将此值

赋给它左边的那个值(这里说的通俗些，坑了勿喷!)最后代码呈上。有需要的可留言，可以找我要代码一起学习。

function GameScene:leftCombineNumber()    if (isPlay == false) then        return true    end        local card    local nextCard    for j = 0,3 do        for i = 0,3 do            card = cardList[i..":"..j]            if(card:getData() ~= 0) then                local k = i+1                while(k < 4) do                    nextCard = cardList[k..":"..j]                    if (nextCard:getData() ~=0) then                        if(card:getData() == nextCard:getData()) then                            card:setNum(card:getData()*2)                            nextCard:setNum(0)                            totalScore = totalScore+card:getData()                        end                         k = 4                        break                    end                       k = k + 1                 end            end        end    end         for j = 0,3 do        for i = 0,3 do            card = cardList[i..":"..j]            if(card:getData() == 0) then                local k = i+1                while(k < 4) do                    nextCard = cardList[k..":"..j]                    if (nextCard:getData() ~=0) then                        card:setNum(nextCard:getData())                        nextCard:setNum(0)                        k = 4                    end                       k = k + 1                 end            end        end    end    self:updateNumber()end

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