確定字符串是否包含所有唯一字符

Question

這是破解編碼面試書

問題 實現一個算法來確定一個字符串是否包含所有唯一的字符。 如果不能使用額外的數據結構怎么辦？

我想知道下面的 if 語句中發生了什么？ 誰能給我解釋一下？

我在評論中留下了我對代碼的理解。如果我錯了，請糾正我

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

這是 else 語句嗎

不，否則代碼中會有一個else 。 但在這種情況下， else是不必要的，因為如果char_set[val]為真，方法的執行會立即停止，因為return false; 操作說明。

我不明白為什么我們最后返回 true ？

因為沒有找到重復項，該方法必須返回 true 以指示該字符串由唯一字符組成。 如果找到重復項，該方法將在

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

Answer 3

我只會使用正則表達式，它只需要一行代碼：

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

分解正則表達式：

• (.)捕獲每個字符
• (?!.*?\\\\2)是對捕獲組的反向引用的負面展望

總之，這些意味着“一個不會再出現的角色”

• (...)*上面的意思是其中的 0-n 個

總而言之，它的意思是“由稍后在字符串中重新出現的字符組成”，即獨特的字符。

Answer 4

沒有任何額外數據結構的單行解決方案：

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