LeeCode 08.07. 无重复字符串的排列组合

忘是亡心i 2022-11-17 11:22 123阅读 0赞

### 文章目录 ###

*  题目
 *  解法
 *  代码

# 题目 #

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl8zODMzNjY1OA_size_16_color_FFFFFF_t_70]

*  [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。][Link 1]

# 解法 #

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl8zODMzNjY1OA_size_16_color_FFFFFF_t_70 1]

# 代码 #

public static void main(String[] args) { 
            System.out.println(JSON.toJSONString(permutation("abc")));
        }
    
        public static String[] permutation(String S) { 
            List<String> list = new ArrayList<>();
            String[] strs = S.split("");
            a("",strs,list);
            return list.toArray(new String[strs.length]);
        }
    
    
    
        public static void a(String str, String[] strs, List<String> list){ 
            if(strs.length==1){ 
                str+=strs[0];
                list.add(str);
            }
            for(int i = 0;i<strs.length;i++){ 
                String str1 = str + strs[i];
                List<String> lastList =  new ArrayList<>(Arrays.asList(strs));
                lastList.remove(strs[i]);
                a(str1,lastList.toArray(new String[strs.length-1]),list);
            }
        }

[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl8zODMzNjY1OA_size_16_color_FFFFFF_t_70]: /images/20221022/fc8c2cd266494d2a8e0e69012cc63718.png
[Link 1]: https://leetcode-cn.com/problems/permutation-i-lcci/
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl8zODMzNjY1OA_size_16_color_FFFFFF_t_70 1]: /images/20221022/6f41c9fc767641169669e1b71eb473d4.png