JS计算每个字符出现的次数

今天在优化代码的时候，需要计算每个字符出现的次数，自己写了一个，但是代码过长，效率不高，百度了很久，没有找到什么合适，在一位好友的帮助下：淡描温情

写下下面代码　

//计算每个字符出现的次数
var t = "stringthatresults", i = 0, len = t.length,
result = [],//输出结果
hash = {};//hash表,用于记录是否重复计算字符
for (i = 0; i < len; i++) {
if (!hash[t.charAt(i)]) {
hash[t.charAt(i)] = true;
//分割当前字符本身，来获取出现次数
result.push(t.charAt(i) + ":" + (t.split(t.charAt(i)).length - 1));
}
}
document.writeln(JSON.stringify(result) + "<br />");

var a = "stringthatresults";
var b = {};
var c = null;
for (var i = 0; i < a.length; i++) {
if (!isNaN(b[a.charAt(i)]++) || (b[a.charAt(i)] = 1)) {
c = b[a.charAt(i)] > c ? a.charAt(i) : c;
}
}
document.writeln(JSON.stringify(b) + "<br />");

result = [];
var count = 0, tmp = "", arr = t.split("");
for (i = 0; i < len; i++) {

//替换当前字符，再通过与原有字符串长度做加减，来获取出现次数
tmp = t.replace(new RegExp(arr[i], "g"), "");
count = t.length - tmp.length;

if (count > 0) {
result.push(arr[i] + ":" + count);
}
t = tmp;
}
document.writeln(JSON.stringify(result));

　　结果

["s:3","t:4","r:2","i:1","n:1","g:1","h:1","a:1","e:1","u:1","l:1"]

{"s":3,"t":4,"r":2,"i":1,"n":1,"g":1,"h":1,"a":1,"e":1,"u":1,"l":1}

["s:3","t:4","r:2","i:1","n:1","g:1","h:1","a:1","e:1","u:1","l:1"]

应该还有更好的思路，希望大家多多提建议。个人偏向第一种写法