BitMask 相关

318. Maximum Product of Word Lengths

Given a string array words, return the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. If no such two words exist, return 0.

 Example 1:

Input: words = ["abcw","baz","foo","bar","xtfn","abcdef"]
Output: 16
Explanation: The two words can be "abcw", "xtfn".

Example 2:

Input: words = ["a","ab","abc","d","cd","bcd","abcd"]
Output: 4
Explanation: The two words can be "ab", "cd".

Example 3:

Input: words = ["a","aa","aaa","aaaa"]
Output: 0
Explanation: No such pair of words. 

Constraints:

• 2 <= words.length <= 1000
• 1 <= words[i].length <= 1000
• words[i] consists only of lowercase English letters.

暴力解法

class Solution {
    public int maxProduct(String[] words) {
        int max = 0;
        for(int i=0;i<words.length-1;i++){
            for(int j=i+1;j<words.length;j++){
                if(!hasCommon(words[i],words[j]))
                    max = Math.max(max,words[i].length()*words[j].length());
            }
        }
        return max;
    }
    private boolean hasCommon(String w1,String w2){
        for(char c : w1.toCharArray()){
            if(w2.indexOf(c)>=0) return true;
        }
        return false;
    }
}

 

巧用bitmask，将 字符串比对 过程从O(len2) 到 O(1)， 最终时间复杂度： O(N*N+L) N为字符串个数，L为字符串长度

class Solution {
    public int maxProduct(String[] words) {
        int max = 0;
        int[] state = new int[words.length];
        for(int i=0;i<words.length;i++){
            state[i] = getState(words[i]);
        }
        for(int i=0;i<words.length-1;i++){
            for(int j=i+1;j<words.length;j++){
                if( (state[i] & state[j]) ==0 )
                    max = Math.max(max,words[i].length()*words[j].length());
            }
        }
        return max;
    }
    private int getState(String str){
        int mask = 0;
        for(char c:str.toCharArray()){
            int pos = c-'a';
            mask |= 1<<(pos);
        }
        return mask;
    }
}

bitmask 基础上增加 map对重复字符串集合进行去重，减少比对次数

class Solution {
    public int maxProduct(String[] words) {
        int max = 0;
        Map<Integer,Integer> map = new HashMap();
        
        for(int i=0;i<words.length;i++){
            int bitValue = getState(words[i]);
            map.put(bitValue, Math.max(map.getOrDefault(bitValue,0), words[i].length()));
        }
        int result = 0;
        for(int i:map.keySet()){
            for(int j:map.keySet()){
                if( i!=j && ( (i & j) == 0))
                result = Math.max(result, map.get(i)*map.get(j));
            }
        }
        return result;
    }
    private int getState(String str){
        int mask = 0;
        for(char c:str.toCharArray()){
            int pos = c-'a';
            mask |= 1<<(pos);
        }
        return mask;
    }
}

 

标签：return,String,int,Explanation,two,BitMask,words,相关
来源： https://www.cnblogs.com/cynrjy/p/16325459.html