69 lines
1.8 KiB
Java
69 lines
1.8 KiB
Java
//有重复字符串的排列组合。编写一种方法，计算某字符串的所有排列组合。
|
||
//
|
||
// 示例1:
|
||
//
|
||
// 输入：S = "qqe"
|
||
// 输出：["eqq","qeq","qqe"]
|
||
//
|
||
//
|
||
// 示例2:
|
||
//
|
||
// 输入：S = "ab"
|
||
// 输出：["ab", "ba"]
|
||
//
|
||
//
|
||
// 提示:
|
||
//
|
||
//
|
||
// 字符都是英文字母。
|
||
// 字符串长度在[1, 9]之间。
|
||
//
|
||
// Related Topics 回溯算法
|
||
// 👍 37 👎 0
|
||
|
||
package leetcode.editor.cn;
|
||
|
||
import java.util.*;
|
||
|
||
//面试题 08.08:有重复字符串的排列组合
|
||
public class PermutationIiLcci {
|
||
public static void main(String[] args) {
|
||
//测试代码
|
||
Solution solution = new PermutationIiLcci().new Solution();
|
||
solution.permutation("ab");
|
||
}
|
||
|
||
//力扣代码
|
||
//leetcode submit region begin(Prohibit modification and deletion)
|
||
class Solution {
|
||
List<String> list;
|
||
|
||
public String[] permutation(String S) {
|
||
list = new ArrayList<>();
|
||
char[] str = S.toCharArray();
|
||
Arrays.sort(str);
|
||
boolean[] visited = new boolean[str.length];
|
||
dfs(str, visited, new StringBuilder());
|
||
return list.toArray(new String[0]);
|
||
}
|
||
|
||
private void dfs(char[] str, boolean visited[], StringBuilder temp) {
|
||
if (temp.length() == str.length) {
|
||
list.add(temp.toString());
|
||
return;
|
||
}
|
||
for (int i = 0; i < str.length; i++) {
|
||
if (visited[i] || i != 0 && !visited[i - 1] && str[i] == str[i - 1]) {
|
||
continue;
|
||
}
|
||
temp.append(str[i]);
|
||
visited[i] = true;
|
||
dfs(str, visited, temp);
|
||
temp.deleteCharAt(temp.length() - 1);
|
||
visited[i] = false;
|
||
}
|
||
}
|
||
}
|
||
//leetcode submit region end(Prohibit modification and deletion)
|
||
|
||
} |