【刷题】Leetcode 1239. Maximum Length of a Concatenated String with Unique Characters

做一道新题。

Example 1:

Input: arr = [“un”,“iq”,“ue”]
Output: 4
Explanation: All possible concatenations are “”,“un”,“iq”,“ue”,“uniq” and “ique”.
Maximum length is 4.
图转自花花酱
用bitmask

class Solution {
    int max = 0;
    public int maxLength(List<String> arr) {      
        List<Integer> list = new LinkedList<Integer>();       
        for(String a: arr){
            int curMask = 0;
            for(int i = 0; i < a.length(); i++){             
                curMask = curMask | (1 << a.charAt(i) -'a');                
            } 
            if(Integer.bitCount(curMask) != a.length())
                    continue;
                list.add(curMask);
        }
        helper(list, 0, 0);
        return max;
    }
    
    public void helper(List<Integer> list, int n, int mask){                       
        int count = Integer.bitCount(mask);
        max = Math.max(count, max);
        for(int i = n; i < list.size(); i++){    
            if((mask & list.get(i)) == 0)
            {
               helper(list, i + 1, mask | list.get(i));
            }                          
        }   
    }
}

用set做也可以

class Solution {
    int max = 0;
    public int maxLength(List<String> arr) {      
        if(arr == null)
            return 0;
        helper(arr, "", 0);
        return max;
    }    
    public void helper(List<String> arr, String path, int n){
        max = Math.max(max, path.length());
        for(int i = n; i < arr.size(); i++){
            if(isUnique(path + arr.get(i))){
                helper(arr, path + arr.get(i), i + 1);
            }
        }
    }
    public boolean isUnique(String path){
        HashSet<Character> set = new HashSet<Character>();
        for(int i = 0; i < path.length(); i++){
            if(set.contains(path.charAt(i)))
                return false;
            set.add(path.charAt(i));
        }
        return true;
    }
  
}

总结

• 先把string转成bitmask
• Integer.bitCount()求1的个数