[leetCode]面试题 08.07. 无重复字符串的排列组合

回溯法

每使用一个字符就往栈中添加一个字符，如果该字符已经使用过则跳过该字符。回溯时从栈中删除字符。

class Solution {
    public String[] permutation(String S) {
        if (S.length() == 0) return new String[]{};
        List<String> ans = new ArrayList<>();
        Deque<Character> path = new ArrayDeque<>();
        boolean[] used = new boolean[S.length()]; 
        dfs(S, 0, used, path, ans);
        return ans.toArray(new String[ans.size()]);
    }

    private void dfs(String s, int depth, boolean[] used, Deque<Character> path, List<String> ans) {
        if (depth == s.length()) {
            StringBuffer buffer = new StringBuffer();
            List<Character> list = new ArrayList(path);
            for (Character c : list) {
                buffer.append(c);
            }
            ans.add(buffer.toString());
        }
        for (int i = 0; i < s.length(); i++) {
            if (used[i]) {
                continue;
            }
            path.addLast(s.charAt(i));
            used[i] = true;
            dfs(s, depth + 1, used, path, ans);
            path.pollLast();
            used[i] = false;
        }
    }
}

回溯 交换法

依次选择每个字符与其左边字符交换，回溯时交换回原来位置

class Solution {
    public String[] permutation(String S) {
        if (S.length() == 0) return new String[]{};
        List<String> ans = new ArrayList<>();
        List<Character> output = new ArrayList<>();
        for (int i = 0; i < S.length(); i++) {
            output.add(S.charAt(i));
        }
        dfs(S.length(), 0, output, ans);
        return ans.toArray(new String[ans.size()]);
    }

    private void dfs(int len, int begin, List<Character> output, List<String> ans) {
        if (begin == len) {
            StringBuffer buffer = new StringBuffer();
            for (Character c : output) {
                buffer.append(c);
            }
            ans.add(buffer.toString());
        }
        for (int i = begin; i < len; i++) {
            Collections.swap(output, i, begin);
            dfs(len, begin + 1, output, ans);
            Collections.swap(output, i, begin);
        }
    }
}