Question

我认为我根本没有做到这一点......我是朝着正确的方向前进的吗？我试图实现循环来检查字符串中是否有每个字母＆＃34; beg＆＃34;匹配数组中的字母。

＆＃34;求＆＃34;是为我的作业提供的文字

＆＃13;

//
// ***(15)  store the number of times the letter "a" appears in the string "beg" in 1st location;
// ***      store the number of times the letter "b" appears in the string "beg" in 2nd location;
// ***      store the number of times the letter "c" appears in the string "beg" in 3rd location;
// ***      store the number of times the letter "d" appears in the string "beg" in 4th location;
// ***      etc.
// ***      show the 26 counts on one line separated by commas in the span block with id="ans15"
//



var alphaNum = [26];
var alphabet = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n"
  "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"
];

for (i = 0; i < 26; i++) {
  alphaNum[i] = 0;
}
for (i = 0; i < beg.length; j++) {
  charNow = beg.substr(i, 1);
  for (j = 0; j < 26; j++) {
    if (charNow == alphabet[j])
      alphaNum = alphaNum[j] + 1;
  }
}


showAlpha = "";
for (i = 0; i < 26; i++) {
  showAlpha = showAlpha + alphabet[i] + ": " + alphaNum[i] + "<br>"
}

ans15.innerHTML = showAlpha;

＆＃13;

Answer 1

beg丢失，在alphabet，缺少逗号，其他一些错误在评论中。

至少它正在运行代码并进行一些小改动。

var beg = 'store the number of times the letter "a" appears in the string "beg" in 1st location;', // declaration missing
    alphaNum = [],          // empty array, not an array with one element 26
    alphabet = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"],
    i, j,           // declaration missing
    charNow,        // declaration missing
    showAlpha = ""; // declaration missing

for (i = 0; i < 26; i++) {
    alphaNum[i] = 0;
}
for (i = 0; i < beg.length; i++) {         // should be i++
    charNow = beg.substr(i, 1);            // could be replaced by charNow = beg[i]
    for (j = 0; j < 26; j++) {
        if (charNow == alphabet[j]) {      // adding some curly brackets
            alphaNum[j] = alphaNum[j] + 1; // index missing
        }
    }
}
for (i = 0; i < 26; i++) {
    showAlpha = showAlpha + alphabet[i] + ": " + alphaNum[i] + "<br>"
}

document.getElementById('ans15').innerHTML = showAlpha; // target missing

<div id="ans15"></div>

存储字母表中每个字母出现在所需位置的字符串中的次数

1 个答案: