【leetcode】318（Medium）Maximum Product of Word Lengths

解题思路

两两对比字符串，记录所有符合条件的字符串组的乘积最大值

提交代码

class Solution {
    public int maxProduct(String[] words) {
    	int res=0,tmp=0;
    	for(int i=0;i<words.length;i++) {
    		for(int j=i+1;j<words.length;j++) {
    			tmp=words[i].length()*words[j].length();
    			if(tmp<res)	continue;
    			if(!hasCommon(words[i],words[j]))
    				res=tmp;
    		}
    	}
    	return res;
    }
    
    private boolean hasCommon(String s1,String s2) {
    	Map<Character,Integer> map1=new HashMap<>();
    	Map<Character,Integer> map2=new HashMap<>();
    	
    	for(int i=0;i<s1.length();i++) {
    		if(!map1.containsKey(s1.charAt(i)))
    			map1.put(s1.charAt(i),1);
    	}
    	
    	for(int i=0;i<s2.length();i++) {
    		if(!map2.containsKey(s2.charAt(i)))
    			map2.put(s2.charAt(i),1);
    	}
    	
    	for(Character key : map1.keySet()) {
    		if(map2.containsKey(key))
    			return true;
    	}
    	return false;
    }
}

运行结果：

优化后的代码

class Solution{
    public int maxProduct(String[] words) {
	if (words == null || words.length == 0) return 0;
	int len=words.length;
        int[] charSet = new int[len];
	for (int i = 0; i < len; i++) {
		String word = words[i];
        for(char c : word.toCharArray())
            charSet[i]|=1<<(c-'a');
	}
	int res = 0;
	for (int i = 0; i < len; i++)
		for (int j = i + 1; j < len; j++) {
			if ((charSet[i] & charSet[j]) == 0 && (words[i].length() * words[j].length() > res))
				res = words[i].length() * words[j].length();
		}
	return res;
}
}

运行结果