求个算法.

Question

var list = [
    {color:"red",    size:"x", count:4},
    {color:"red",    size:"l", count:3},
    {color:"yellow", size:"l", count:2},
    {color:"red",    size:"x", count:1},
    {color:"blue",   size:"x", count:3},
    {color:"yellow", size:"l", count:2}
]//希望返回如下数组var return = [
    {color:"red",    size:"x", count:5},
    {color:"red",    size:"l", count:3},
    {color:"yellow", size:"l", count:4},
    {color:"blue",   size:"x", count:3},
]//即 color和size相同的.合并为同一个对象,更改count,无需排序,谢谢.

Answer 1

写了一个，你看看：var list = [
    {color:"red",    size:"x", count:4},
    {color:"red",    size:"l", count:3},
    {color:"yellow", size:"l", count:2},
    {color:"red",    size:"x", count:1},
    {color:"blue",   size:"x", count:3},
    {color:"yellow", size:"l", count:2}
];
var res=[];
for(var i=0,temp={};i<list.length;i++) {
var cur=list[i];
var t=temp[cur.color+"|"+cur.size];
if(typeof(t)!="undefined" && res[t[0]].color==cur.color && res[t[0]].size==cur.size) {
res[t[1]].count+=cur.count;
}else {
res.push(cur);
temp[cur.color+"|"+cur.size]=[i,res.length-1];
}
}//输出测试
var arr=[];
for(var i=0;i<res.length;i++) {
arr.push(res[i].color+"|"+res[i].size+"|"+res[i].count);
}
document.write(arr.join("<br/>"));

Answer 2

<script type="text/javascript">
<!--
var list = [
    {color:"red",    size:"x", count:4},
    {color:"red",    size:"l", count:3},
    {color:"yellow", size:"l", count:2},
    {color:"red",    size:"x", count:1},
    {color:"blue",   size:"x", count:3},
    {color:"yellow", size:"l", count:2}
];var count = 0, arr = [];
for(var j = 0; j < list.length; j++){
list[j].index = j;
}
function By(color, size){
return function(o, p){
var a, b, c, d;
if(typeof o === 'object' && typeof p === 'object' && o && p){
 a =  o[color]; b = p[color]; c = o[size]; d = p[size];
 if((a == b) && (c == d)){
 o['count'] += p['count'];
 arr.push(p.index);
 }
}
return 0;
}
}list.sort(By('color', 'size'))
for(var i = 0; i < arr.length; i++){
list.splice(arr[i], 1);
}
for(var k = 0; k < list.length; k++){
document.write(list[k].color + "--->" + list[k].size + "--->" + list[k].count + "<br/>");
}//希望返回如下数组/*var return = [
    {color:"red",    size:"x", count:5},
    {color:"red",    size:"l", count:3},
    {color:"yellow", size:"l", count:4},
    {color:"blue",   size:"x", count:3},
]*///-->
</script>

Answer 3

<script>var list = [
    {color:"red",    size:"x", count:4},
    {color:"red",    size:"l", count:3},
    {color:"yellow", size:"l", count:2},
    {color:"red",    size:"x", count:1},
    {color:"blue",   size:"x", count:3},
    {color:"yellow", size:"l", count:2}
]var hash = {},result=[];
for(var i=0,n=list.length;i<n;i++){
var l=list[i],name=l.color+"_"+l.size;
if(hash[name]){
hash[name]+=l.count;
}else{
hash[name]=l.count;
}
}
for(var name in hash){
var n=name.split("_")
result.push({color:n[0], size:n[1], count:hash[name]});
}
for(var i=0,n=result.length;i<n;i++){
var r=result[i];
document.write("{color:"+r.color+", size:"+r.size+", count:"+r.count+"}<br>")
}</script>

Answer 4

反向循环嵌套合并，算不算优雅？！L@_@K
var list = [
    {color:"red",    size:"x", count:4},
    {color:"red",    size:"l", count:3},
    {color:"yellow", size:"l", count:2},
    {color:"red",    size:"x", count:1},
    {color:"blue",   size:"x", count:3},
    {color:"yellow", size:"l", count:2}
];var cur, sibling;for (var i=list.length-1; i >= 0; i--)
{
    cur = list[i];    for (var j=i-1; j >= 0; j--)
    {
        sibling = list[j];        if (cur.color == sibling.color 
            && cur.size == sibling.size)
        {
            sibling.count += cur.count;
            list.splice(i, 1);
            break;
        }
    }
}alert(list.length);
for (var i=0; i<list.length; i++)
{
    document.write("color: ", list[i].color
                  ,", size: ", list[i].size
                  ,", count:", list[i].count
                  ,"<br />");
}

Answer 5

晕，俺修改了原数组的元素，lz 希望这样么？！不过这引出一个新问题，JS 数组如何实现深度复制？！

Answer 6

递归一下就行啦
参考那个deepextend

Answer 7

哈，多谢顺便问下，cloudgamer 怎么翻译？云游者？！

Answer 8

to #4 sohighthesky 谢谢你的回复,你的代码我看懂了. 不过一个判断里面好像有点错误,  if(typeof(t)!="undefined" && res[t[0]].color==cur.color && res[t[0]].size==cur.size)这里可能得用 res[t[1]].color  res[t[1]].size 来对比,不知道是不是我搞错了.

Answer 9

是的，疏忽了，我原本是想这么写的：list[t[0]].color==cur.color && list[t[1]].size==cur.size

Answer 10

既然这样后面的值就不用数组了：var res=[];
for(var i=0,temp={};i<list.length;i++) {
var cur=list[i];
var t=temp[cur.color+"|"+cur.size];
if(typeof(t)!="undefined" && res[t].color==cur.color && res[t].size==cur.size) {
res[t].count+=cur.count;
}else {
res.push(cur);
temp[cur.color+"|"+cur.size]=res.length-1;
}
}