面试题 08.07. 无重复字符串的排列组合

无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，字符串每个字符均不相同。

示例1:

输入：S = "qwe"
输出：["qwe", "qew", "wqe", "weq", "ewq", "eqw"]

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。

1.回溯
python:

class Solution:
    def permutation(self, S: str) -> List[str]:
        res = []
        A = list(S)
        flag = [0 for i in range(len(A))]
        def backtrace(A, arr):
            if len(arr) == len(A):
                res.append(''.join(arr)) 
            for i in range(len(A)):
                if not flag[i]:
                    arr.append(A[i])
                    flag[i] = 1
                    backtrace(A, arr)
                    arr.pop()
                    flag[i] = 0
        backtrace(A, [])
        return res

Java:

class Solution {
    public String[] permutation(String S) {
        List<String> result = new ArrayList<String>();
        boolean[] flag = new boolean[S.length()];
        backtrack(result, flag, S, "");
        return result.toArray(new String[result.size()]);

    }

    public void backtrack(List<String> a, boolean[] flag, String s, String t)
    {
        if(s.length() == t.length())
        {
            a.add(t);
        }
        for(int i = 0; i < s.length(); i++)
        {
            if(!flag[i])
            {
                flag[i] = true;
                backtrack(a, flag, s, t+s.charAt(i));
                flag[i] = false;
            }
        }
    }
}

2.交换法
Python:

class Solution:
    def permutation(self, S: str) -> List[str]:
        res = []
        a = list(S)
        def swap(arr, j):
            if j == len(arr):
                res.append(''.join(arr))
                return
            else:
                for i in range(j, len(arr)):
                    arr[i], arr[j] = arr[j], arr[i]
                    swap(arr, j+1)
                    arr[i], arr[j] = arr[j], arr[i]
        swap(a, 0)
        return res

Java:

class Solution {
    public String[] permutation(String S) {
        List<String> result = new ArrayList<String>();
        char[] ch = new char[S.length()];
        for(int i = 0; i < S.length(); i++)
        {
            ch[i] = S.charAt(i);
        }
        boolean[] flag = new boolean[S.length()];
        backtrack(result, ch, 0);
        return result.toArray(new String[result.size()]);
    }

    public void backtrack(List<String> a, char[] ch, int p)
    {
        if(p == ch.length)
        {
            a.add(String.valueOf(ch));
        }
        for(int i = p; i < ch.length; i++)
        {
            swap(ch, i, p);
            backtrack(a, ch, p+1);
            swap(ch, i, p);
        }
    }
    public void swap(char[] ch, int i, int j)
    {
        char t = ch[i];
        ch[i] = ch[j];
        ch[j] = t;
    }
}