JZ-027-String arrangement

String arrangement

Title description

Enter a character string and print out all the characters in the character string in lexicographic order. For example, input the string abc, and print out all the strings abc, acb, bac, bca, cab and cba that can be arranged by the characters a, b, and c in lexicographic order.

topic link : string arrangement

Code

import java.util.ArrayList;
import java.util.Arrays;

/**
 * 标题：字符串的排列
 * 题目描述
 * 输入一个字符串,按字典序打印出该字符串中字符的所有排列。例如输入字符串abc,
 * 则按字典序打印出由字符a,b,c所能排列出来的所有字符串abc,acb,bac,bca,cab和cba。
 * 题目链接：
 * https://www.nowcoder.com/practice/fe6b651b66ae47d7acce78ffdd9a96c7?tpId=13&&tqId=11180&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking
 */
public class Jz27 {
  private ArrayList<String> ret = new ArrayList<>();

  public ArrayList<String> permutation(String str) {
    if (str.length() == 0) {
      return ret;
    }
    char[] chars = str.toCharArray();
    Arrays.sort(chars);
    backtracking(chars, new boolean[chars.length], new StringBuilder());
    return ret;
  }

  private void backtracking(char[] chars, boolean[] hasUsed, StringBuilder s) {
    if (s.length() == chars.length) {
      ret.add(s.toString());
      return;
    }
    for (int i = 0; i < chars.length; i++) {
      if (hasUsed[i]) {
        continue;
      }
      // 保证不重复
      if (i != 0 && chars[i] == chars[i - 1] && hasUsed[i - 1]) {
        continue;
      }
      hasUsed[i] = true;
      s.append(chars[i]);
      backtracking(chars, hasUsed, s);
      s.deleteCharAt(s.length() - 1);
      hasUsed[i] = false;
    }
  }

  public static void main(String[] args) {
    Jz27 jz27 = new Jz27();
    ArrayList<String> result = jz27.permutation("abcdee");
    System.out.println("size: " + result.size());
    for (int i = 0; i < result.size(); i++) {
      System.out.println(result.get(i));
    }
  }
}

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