javascript冒泡排序小结

冒泡排序示例，双向冒泡排序与对双向冒泡排序稍微的改进的可视化效果。

代码很简单，不知道有木有未知bug。
大神请勿吐槽

冒泡排序示例

var ls=[ 98,13,6,25,38,36,30,44,38,80,61,28,47,34,95,18,85,58,89,85,42,61,74,35,13,14,80,7,10,44,10,47,13,11,52,25,24,48,34,12,88,80,33,8,80,45,64,52,79,77 ];

for(var i=0;i<ls.length;i++){
      for(var j=i+1;j<ls.length;j++){
        if(ls[i]>ls[j]){
          ls[i]=ls[i]+ls[j];
          ls[j]=ls[i]-ls[j];
          ls[i]=ls[i]-ls[j];
        } 
      }
    }

双向冒泡排序示例

var ls=[ 6,13,98,25,38,36,30,44,38,80,61,28,47,34,95,18,85,58,89,85,42,61,74,35,13,14,80,7,10,44,10,47,13,11,52,25,24,48,34,12,88,80,33,8,80,45,64,52,79,77 ];

for(var i=0;i<ls.length;i++){
      for(var j=i+1;j<ls.length-i;j++){
        if(ls[lent-1-i]<ls[lent-j]){
          ls[lent-1-i]=ls[lent-1-i]+ls[lent-j];
          ls[lent-j]=ls[lent-1-i]-ls[lent-j];
          ls[lent-1-i]=ls[lent-1-i]-ls[lent-j];
        }//后面的比较
        if(ls[i]>ls[j]){
          ls[i]=ls[i]+ls[j];
          ls[j]=ls[i]-ls[j];
          ls[i]=ls[i]-ls[j];
        }//前面的比较
      }
    }

双向冒泡排序稍微改进示例

var ls=[ 98,13,6,25,38,36,30,44,38,80,61,28,47,34,95,18,85,58,89,85,42,61,74,35,13,14,80,7,10,44,10,47,13,11,52,25,24,48,34,12,88,80,33,8,80,45,64,52,79,77 ];

var lent=ls.length;
for(var i=0;i<ls.length;i++){
      for(var j=i*2;j<ls.length-2*i;j++){
        if(ls[i*2]>ls[j+1]){
          ls[i*2]=ls[i*2]+ls[j+1];
          ls[j+1]=ls[i*2]-ls[j+1];
          ls[i*2]=ls[i*2]-ls[j+1];
        }//保持内层第一个数为循环最小

        if(ls[lent-i*2-1]<ls[lent-j-1]){
          ls[lent-i*2-1]=ls[lent-i*2-1]+ls[lent-j-1];
          ls[lent-j-1]=ls[lent-i*2-1]-ls[lent-j-1];
          ls[lent-i*2-1]=ls[lent-i*2-1]-ls[lent-j-1];
        }////保持内层倒数第一个数为循环最大

          if(ls[lent-2-i*2]<ls[lent-j-1]){
            ls[lent-2-i*2]=ls[lent-2-i*2]+ls[lent-j-1];
            ls[lent-j-1]=ls[lent-2-i*2]-ls[lent-j-1];
            ls[lent-2-i*2]=ls[lent-2-i*2]-ls[lent-j-1];
          }//倒数上一个

          if(ls[i*2+1]>ls[j+1]){
            ls[i*2+1]=ls[i*2+1]+ls[j+1];
            ls[j+1]=ls[i*2+1]-ls[j+1];
            ls[i*2+1]=ls[i*2+1]-ls[j+1];
          }//下一个
      }
}