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面试题 08.08. 有重复字符串的排列组合 ​

题目 ​

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"]
 输入：S = "qqe"
 输出：["eqq","qeq","qqe"]

示例2:

 输入：S = "ab"
 输出：["ab", "ba"]
 输入：S = "ab"
 输出：["ab", "ba"]

提示:

  1. 字符都是英文字母。
  2. 字符串长度在[1, 9]之间。

题解 ​

java
public String[] permutation(String S) {
    char[] chars = S.toCharArray();
    List<String> result = new ArrayList<>();
    if (chars.length == 0) {
        return new String[0];
    }
    // 通过长度缓存已经使用过的前缀
    Map<Integer, Set<String>> visited = new HashMap<>(9);
    Consumer<Integer> recursion = new Consumer<Integer>() {
        @Override
        public void accept(Integer index) {
            if (index == chars.length) {
                result.add(new String(chars));
                return;
            }

            for (int i = index; i < chars.length; i++) {
                // 选定的字符移动到索引位置 并交换两个位置的字符
                char temp = chars[index];
                chars[index] = chars[i];
                chars[i] = temp;
                String prefix = new String(chars, 0, index + 1);
                // 若前缀未访问
                if (!visited.getOrDefault(index, new HashSet<>()).contains(prefix)) {
                    visited.merge(index, new HashSet<>(Collections.singletonList(prefix)), (o, n) -> {
                        o.addAll(n);

                        return o;
                    });
                    this.accept(index + 1);
                }
                // 回溯
                chars[i] = chars[index];
                chars[index] = temp;
            }
        }
    };
    recursion.accept(0);

    return result.toArray(new String[0]);
}
public String[] permutation(String S) {
    char[] chars = S.toCharArray();
    List<String> result = new ArrayList<>();
    if (chars.length == 0) {
        return new String[0];
    }
    // 通过长度缓存已经使用过的前缀
    Map<Integer, Set<String>> visited = new HashMap<>(9);
    Consumer<Integer> recursion = new Consumer<Integer>() {
        @Override
        public void accept(Integer index) {
            if (index == chars.length) {
                result.add(new String(chars));
                return;
            }

            for (int i = index; i < chars.length; i++) {
                // 选定的字符移动到索引位置 并交换两个位置的字符
                char temp = chars[index];
                chars[index] = chars[i];
                chars[i] = temp;
                String prefix = new String(chars, 0, index + 1);
                // 若前缀未访问
                if (!visited.getOrDefault(index, new HashSet<>()).contains(prefix)) {
                    visited.merge(index, new HashSet<>(Collections.singletonList(prefix)), (o, n) -> {
                        o.addAll(n);

                        return o;
                    });
                    this.accept(index + 1);
                }
                // 回溯
                chars[i] = chars[index];
                chars[index] = temp;
            }
        }
    };
    recursion.accept(0);

    return result.toArray(new String[0]);
}