删除字符串中重复字符的功能

Question

下面的代码试图删除字符串中所有重复的字符。我不确定代码是否正确。有人可以帮助我处理代码吗（即，字符匹配时实际发生了什么）？

public static void removeDuplicates(char[] str) {
  if (str == null) return;
  int len = str.length;
  if (len < 2) return;
  int tail = 1;
  for (int i = 1; i < len; ++i) {
    int j;
    for (j = 0; j < tail; ++j) {
      if (str[i] == str[j]) break;
    }
    if (j == tail) {
      str[tail] = str[i];
      ++tail;
    }
  }
  str[tail] = 0;
}

Answer 1

该功能对我来说很好。我已经写了内联评论。希望能帮助到你：

// function takes a char array as input.
// modifies it to remove duplicates and adds a 0 to mark the end
// of the unique chars in the array.
public static void removeDuplicates(char[] str) {
  if (str == null) return; // if the array does not exist..nothing to do return.
  int len = str.length; // get the array length.
  if (len < 2) return; // if its less than 2..can't have duplicates..return.
  int tail = 1; // number of unique char in the array.
  // start at 2nd char and go till the end of the array.
  for (int i = 1; i < len; ++i) { 
    int j;
    // for every char in outer loop check if that char is already seen.
    // char in [0,tail) are all unique.
    for (j = 0; j < tail; ++j) {
      if (str[i] == str[j]) break; // break if we find duplicate.
    }
    // if j reachs tail..we did not break, which implies this char at pos i
    // is not a duplicate. So we need to add it our "unique char list"
    // we add it to the end, that is at pos tail.
    if (j == tail) {
      str[tail] = str[i]; // add
      ++tail; // increment tail...[0,tail) is still "unique char list"
    }
  }
  str[tail] = 0; // add a 0 at the end to mark the end of the unique char.
}