41 lines
1.3 KiB
Java
41 lines
1.3 KiB
Java
# 38. 字符串的排列
|
||
|
||
[NowCoder](https://www.nowcoder.com/practice/fe6b651b66ae47d7acce78ffdd9a96c7?tpId=13&tqId=11180&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||
|
||
## 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。
|
||
|
||
## 解题思路
|
||
|
||
```java
|
||
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;
|
||
if (i != 0 && chars[i] == chars[i - 1] && !hasUsed[i - 1]) /* 保证不重复 */
|
||
continue;
|
||
hasUsed[i] = true;
|
||
s.append(chars[i]);
|
||
backtracking(chars, hasUsed, s);
|
||
s.deleteCharAt(s.length() - 1);
|
||
hasUsed[i] = false;
|
||
}
|
||
}
|
||
```
|