返回字典序中首先出现的字符串的长度

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英文:

Return the length of the String that comes first lexicographically

问题

以下是翻译好的内容:

public static int problem4(String s, String t) {
    for (int i = 0; i < s.length() && i < t.length(); i++) {
        if ((int) s.charAt(i) == (int) t.charAt(i)) {
            continue;
        } else {
            return (int) s.length() - (int) t.length();
        }
    }

    if (s.length() < t.length()) {
        return (s.length() - t.length());
    } else if (s.length() > t.length()) {
        return (s.length() - t.length());
    } else {
        return 0;
    }
}
英文:

I am trying pass two strings to a function and I want to return the length of the string that comes first lexicographically.

This is what I have tried so far:

public static int problem4(String s, String t) {
    for (int i = 0; i &lt; s.length() &amp;&amp;
            i &lt; t.length(); i++) {
        if ((int) s.charAt(i) ==
                (int) t.charAt(i)) {
            continue;
        } else {
            return (int) s.length() -
                    (int) t.length();
        }
    }

    if (s.length() &lt; t.length()) {
        return (s.length() - t.length());
    } else if (s.length() &gt; t.length()) {
        return (s.length() - t.length());
    }

    // If none of the above conditions is true, 
    // it implies both the strings are equal 
    else {
        return 0;
    }
}

答案1

得分: 2

你可以将它们放入一个数组中,然后使用 Arrays.sort()。然后像这样检索第一个项目:

public static int problem4(String s, String t) {
    String[] items = new String[2];
    items[0] = s;
    items[1] = t;
    Arrays.sort(items);
    return items[0].length();
}

或者你可以使用 .compareTo 方法,像这样:

public static int problem4(String s, String t) {
    return s.compareTo(t) > 0 ? t.length() : s.length();
}
英文:

You can set them into an array, and use Arrays.sort(). Then retrieve the first item like so:

public static int problem4(String s, String t) {
    String[] items = new String[2];
    items[0]=s;
    items[1]=t;
    items.sort();
    return items[0].length();
    
}

Or you can use the .compareTo method like so:

public static int problem4(String s, String t) {
    return s.compareTo(t) &gt; 0 ? t.length() : s.length();
 }

答案2

得分: 0

以下是翻译好的代码部分:

public static int problem4(String s, String t){
    if (s.compareTo(t) > 0)
        System.out.println(t.length());
        return t.length();
    else 
        System.out.println(s.length());
        return s.length();
}

注意:原始代码中的 problem("a", "b"); 似乎有一个小错误,应该是 problem4("a", "b"); 才正确。

英文:

Something to this effect should work I think.

public static int problem4(String s, String t){
    if (s.compareTo(t)&gt;0)
        System.out.println(t.length());
        return t.length();
    else 
        System.out.println(s.length());
        return s.length();
}

problem(&quot;a&quot;, &quot;b&quot;);

huangapple
  • 本文由 发表于 2020年10月6日 20:38:44
  • 转载请务必保留本文链接:https://go.coder-hub.com/64225973.html
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