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字符串中字母的出现次数

Open Sunny-117 opened this issue 3 years ago • 7 comments

Sunny-117 avatar Nov 03 '22 08:11 Sunny-117

const str = "ABCabc123"
let res = 0
for (const c of str)
    if ((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z'))
        res++
console.log(res)  // 6

GlintonLiao avatar Jan 18 '23 00:01 GlintonLiao

function count(str){
    let res = 0;
    for(let i = 0; i < str.length; i++){
        if(str[i] >= 'a' && str[i] <= 'z' || (str[i] >= 'A' && str[i] <= 'Z')) res++;
    }
    return res;
}
console.log(count("ABCabc123"));

如果是要统计每个字母的次数,可以用map

bearki99 avatar Feb 13 '23 02:02 bearki99

const reg = /[^a-zA-Z]/g;
const str = "ABCabc123";
console.log(str.replaceAll(reg, '').length)

fencer-yd avatar Jun 01 '23 09:06 fencer-yd

const str = "ABCabc123";
let res = str.split("").reduce((pre, cur) => {
  if (!pre[cur]) pre[cur] = 1;
  else pre[cur]++;
  return pre;
}, {});
console.log(res);

Banks1y avatar Jun 18 '23 07:06 Banks1y

function countCharacters(str) {
    let charCount = {};
    for(let i = 0; i < str.length; i++) {
        let char = str[i];
        if(char.match(/[a-zA-Z]/i)) { // 只考虑字母
            if(charCount[char]) {
                charCount[char]++;
            } else {
                charCount[char] = 1;
            }
        }
    }
    return charCount;
}

kangkang123269 avatar Sep 11 '23 09:09 kangkang123269

function strNumber(str) {
  let reg = /[a-zA-Z]/;
  let map = new Map();
  for (let i = 0; i < str.length; i++) {
    if (reg.test(str[i])) {
      if (map.has(str[i])) {
        let num = map.get(str[i]);
        map.set(str[i], num + 1);
      } else {
        map.set(str[i], 1);
      }
    }
  }
  console.log(map);
}

strNumber("abcdeabdasjklhsafhgkjsdnioqwjldask123276543");

dizao006 avatar Oct 09 '24 06:10 dizao006

const a = 'abcdeabdasjklhsafhgkjsdnioqwjldask123276543';

function getCount(a) {
  let count = 0;
  let reg = /[a-zA-Z]/;
  console.log(a.length);
  for(var i = 0; i < a.length; i++) {
    if(reg.test(a[i])) {
      count++;
    }
  }
  return count;
}

console.log(getCount(a));

Windseek avatar Nov 16 '24 04:11 Windseek