我正在弄乱第一个问题:Reduce duplicate characters to a desired minimum我正在寻找比我想出的更优雅的答案。它通过了测试,但很想知道其他解决方案。样本测试是:
reduceString('aaaabbbb', 2) 'aabb'
reduceString('xaaabbbb', 2) 'xaabb'
reduceString('aaaabbbb', 1) 'ab'
reduceString('aaxxxaabbbb', 2) 'aaxxaabb'
和我的解决方案(通过这些测试):
reduceString = function(str, amount) {
var count = 0;
var result = '';
for (var i = 0; i < str.length; i++) {
if (str[i] === str[i+1]) {
count++;
if (count < amount) {
result += str[i];
}
} else {
count = 0;
result += str[i];
}
};
return result;
}
答案 0 :(得分:5)
只需使用正则表达式。
var reduceString = function (str, amount) {
var re = new RegExp("(.)(?=\\1{" + amount + "})","g");
return str.replace(re, "");
}
答案 1 :(得分:1)
我想我的最佳解决方案就像
var str = "axxxaabbbbcaaxxxaab",
redStr = (s,n) => s.replace(/(\w)\1+/g,"$1".repeat(n));
console.log(redStr(str,2));
答案 2 :(得分:0)
我试图让它尽可能短:
reduceString = function(str, amount) {
var finalString = '', cL = '', counter;
str.split('').forEach(function(i){
if (i !== cL) counter = 0;
counter++;
cL = i;
if (counter <= amount ) finalString = finalString + i;
});
return finalString;
}
答案 3 :(得分:0)
您可以使用reg表达式。在javascript中测试。
它如何运作:
(.) //match any character
\1 //if it follow by the same character
+{2 //more than 1 times
/g //global
$1 //is 1 time by $1$1 is 2 times
reduceString('aaaabbbb', 2)
reduceString('xaaabbbb', 2)
reduceString('aaaabbbb', 1)
reduceString('aaxxxaabbbb', 2)
function reduceString(txt,num)
{
var canRepeat=['$1'];
for (i=1;i<num;i++)
{
canRepeat.push('$1')
}
canRepeat = canRepeat.join('');
console.log(txt.replace(/(.)\1{2,}/g, canRepeat))
}
答案 4 :(得分:0)
使用正则表达式:
var reduceString = function(str, amount) {
var x = [ ...new Set(str) ];
for (var c of x){
var rex = new RegExp(c + '{'+amount+',}','g');
str = str.replace(rex,string(c,amount));
}
return str;
};
var string = function(c,amount){
for(var i=0,s="";i<amount;i++)s+=c;
return s;
};