Question

这是破解编码面试手册

问题实现一种算法来确定字符串是否具有所有唯一字符。如果你不能使用其他数据结构吗？

我想知道下面的if语句中发生了什么？有人可以向我解释一下吗？

我在评论中留下了对代码的理解。如果我错了，请纠正我

public class Uniquechar2 {

    public static boolean isUniqueChars2(String str) {
         // Create a new boolean array of 256 characters to account for basic a cii and extended ascii characters
         boolean[] charSet = new boolean[256];

         //iterate through the array
         for (int i = 0; i < str.length(); i++) {

             // Assign the value of current value of the iterator i to int variable val.So if we are looping through "hello"  at i = 0 the int value of 'h' will be assigned to val.Is that correct?

             int val = str.charAt(i);

             // Continuing from the example of loping throughout the string "hello" the if statement will see if 'h' is in charSet and since it will be there it will return false /is that what is happening?

             if (charSet[val]) {
                 return false;
             }
             // Is this the else statement? true will be assigned to charSet[h] in this case       
             charSet[val] = true;
         }
         // I dont understand why we are returning true at the end ?
         return true;
    }

Answer 1

public static boolean isUniqueChars2(String str) {
     // Create a new boolean array of 256 characters to account for basic ascii and extended ascii characters
     boolean[] char_set = new boolean[256];

     // Iterate through the string we are testing
     for (int i = 0; i < str.length(); i++) {

         // Get the numerical (ascii) value of the character in the `str` at position `i`.
         int val = str.charAt(i);

         // If char_set[val] has been set, that means that this character was already present in the string. (so in string 'hello' this would be true for the second 'l')
         if (char_set[val]) {
             return false;
         }
         // If the character hasn't been encountered yet (otherwise we would have returned false above), then mark this particular character as present in the string
         char_set[val] = true;
     }
     // If the function hasn't returned false after going through the entire string that means that each character is unique - thus returning true
     return true;
}

Answer 2

这是else语句

不，否则代码中会有else。但在这种情况下，else是不必要的，因为如果char_set[val]为真，则由于return false;指令，方法的执行会立即停止。

我不明白为什么我们最后会回归真实？

因为由于未找到重复项，因此该方法必须返回true以指示该字符串由唯一字符组成。如果找到重复，则该方法已经在

中返回

if (char_set[val]) {
    return false;
}

Answer 3

我只使用正则表达式，它只需要一行代码：

public static boolean isUniqueChars(String str) {
    return str.matches("((.)(?!.*?\\2))*");
}

打破正则表达式：

(.)捕获每个角色
(?!.*?\\2)对于捕获的组

这些意味着＆＃34;一个不会再出现的角色＆＃34;

(...)*表示0-n

总而言之，它意味着＆＃34;由字符组成，这些字符稍后会在字符串＆＃34;中重新出现，即唯一字符。

Answer 4

没有任何额外数据结构的单行解决方案：

str.chars().distinct().count() == (int)str.length();

确定字符串是否包含所有唯一字符

4 个答案: