codewars算法题-求两数组元素长度差值极大值

来源：互联网发布：linux ip设置编辑：程序博客网时间：2024/05/07 02:27

算法要求

You are given two arrays a1 and a2 of strings. Each string is composed with letters from a to z. Let x be any string in the first array and y be any string in the second array. Find max(abs(length(x) − length(y))), If a1 or a2 are empty return -1 in each language except in Haskell where you will return Nothing.

总体思路就是先转换数组，再求极值。

方法一：
sort()排序

function mxdiflg(a1, a2) {    function arrSortByLen(a, b){      if(a.length<=b.length){        return -1;      }      if(a.length>b.length){        return 1;      }    }    if(a1.length==0||a2.length==0){      return -1;    }    a1 = a1.sort(arrSortByLen);    a2 = a2.sort(arrSortByLen);    var a1Len = a1.length,        a2Len = a2.length,        firstData = a2[a2Len-1].length-a1[0].length,        seconData = a1[a1Len-1].length-a2[0].length;    return Math.max(firstData, seconData);}

方法二：
加入数组的map()方法

function mxdiflg(a1, a2) {    function sortByLen(ele, index){      return this[index] = ele.length;    }    if(a1.length ==0 || a2.length ==0){      return -1;    }else {      a1    = a1.map(sortByLen);      a2    = a2.map(sortByLen);      var a1Max = Math.max.apply(null, a1),          a1Min = Math.min.apply(null, a1),          a2Max = Math.max.apply(null, a2),          a2Min = Math.min.apply(null, a2);      return Math.max(Math.abs(a1Max-a2Min), Math.abs(a2Max-a1Min));    }}

0 0