求高效实现一个算法
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悬赏园豆：10
[已解决问题]
解决于 2016-11-09 09:36
有四种颜色不一的球，每个球都有若个，都标有不同大小的数字，随机排序，求算法实现找到四种颜色上面数字最小的四个球
最佳答案
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var balls=[];
//fill some data
balls.push({color:"red",number:1});
balls.push({color:"blue",number:4});
balls.push({color:"yellow",number:10});
balls.push({color:"green",number:4});
balls.push({color:"red",number:3});
balls.push({color:"blue",number:5});
balls.push({color:"yellow",number:8});
balls.push({color:"green",number:2});
balls.push({color:"red",number:4});
balls.push({color:"blue",number:8});
balls.push({color:"yellow",number:7});
balls.push({color:"green",number:2});
balls.push({color:"red",number:6});
balls.push({color:"blue",number:3});
balls.push({color:"yellow",number:1});
balls.push({color:"green",number:14});
var redball,blueball,greenball,yellowball;
for(var i=0;i<balls.length;i++){
switch(balls[i].color){
case "red":
if(redball==null)
redball=balls[i];
else{
if(redball.number>balls[i].number)
redball=balls[i];
}
break;
case "blue":
if(blueball==null)
blueball=balls[i];
else{
if(blueball.number>balls[i].number)
blueball=balls[i];
}
break;
case "yellow":
if(yellowball==null)
yellowball=balls[i];
else{
if(yellowball.number>balls[i].number)
yellowball=balls[i];
}
break;
case "green":
if(greenball==null)
greenball=balls[i];
else{
if(greenball.number>balls[i].number)
greenball=balls[i];
}
break;
}
}
result:
Object {color: "red", number: 1}
Object {color: "green", number: 2}
Object {color: "yellow", number: 1}
Object {color: "blue", number: 3}
收获园豆：10
其他回答(4)
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O(n)吧
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无序的遍历一次就行了.4个变量记每种颜色最小球.每个都对比一次.
所有查找算法都要基于有序的数据
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先判颜色，再判大小就可以了，最好是两次遍历，如果想省市，网上很多数组提取最小数。
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你题目是不是记错了？这几种颜色并没有增加难度，你就看成是单色的找最小O(n)不就完了么？
