Palindrome Permutation II


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    Click here to see the full article post


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    class Solution(object):
        def permutePalindromesDetail(self, compressed, odd_char, result, tmp):
            if not compressed:
                palindrome = ''.join(tmp+[odd_char]+tmp[::-1])
                result.append(palindrome)
                return
            pre = ''
            for i in range(len(compressed)):
                if pre != compressed[i]:
                    tmp.append(compressed[i])
                    self.permutePalindromesDetail(compressed[:i]+compressed[i+1:], odd_char, result, tmp)
                    tmp.pop()
                pre = compressed[i]
    
        def generatePalindromes(self, s):
            word_count = collections.Counter(s)
            odd_char, odd_elements, compressed, result = '', 0, [], []
            for k,v in word_count.items():
                if v % 2 == 1:
                    odd_elements += 1
                    odd_char = k
                for _ in range(v / 2):
                    compressed.append(k)
            if odd_elements > 1:
                return []
    
            result = []
            tmp = []
            compressed.sort()
            self.permutePalindromesDetail(compressed, odd_char, result, tmp)
            return result
    

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    A

    My easy Java solution

    class Solution {
        public List<String> generatePalindromes(String s) {
            List<String> res = new ArrayList<>();
            int[] table = new int[128];
            int cnt = 0;
            for(int i=0;i<s.length();i++){
                table[s.charAt(i)]++;
                cnt = (table[s.charAt(i)] % 2 == 0) ? cnt - 1 : cnt + 1;
            }
            if(s.equals("") || cnt > 1){
                return res;
            }
            String temp = "";
            for(int i=0;i<128 && cnt == 1;i++){
                if(table[i] % 2 == 1){
                    temp += (char) i;
                    break;
                }
            }
            backtrackHelper(table,temp,res,s.length());
            
            return res;
        }
        
        public void backtrackHelper(int[] table,String curr,List<String> res,int l){
            if(curr.length() == l){
                res.add(curr);
                return;
            }
            for(int i=0;i<128;i++){
                if(table[i] > 1){
                    table[i] -= 2;
                    char c = (char) i;
                    String temp = c + curr + c;
                    backtrackHelper(table,temp,res,l);
                    table[i] += 2;
                }
            }
        }
    }
    

  • 0
    J

    Since the Set < String > set = new HashSet < > (); need to store all the permutation before return, so Space Complexity is at most O((n/2+1)!), which is the same as time complexity, right?


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