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- Maximum Product of Word Lengths
Given a string array words, find the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0.
Example 1:
Input: [“abcw”,“baz”,“foo”,“bar”,“xtfn”,“abcdef”]
Output: 16
Explanation: The two words can be “abcw”, “xtfn”.
Example 2:
Input: [“a”,“ab”,“abc”,“d”,“cd”,“bcd”,“abcd”]
Output: 4
Explanation: The two words can be “ab”, “cd”.
Example 3:
Input: [“a”,“aa”,“aaa”,“aaaa”]
Output: 0
Explanation: No such pair of words.
思路
如果用HashMap,需要比较字符次数比较多,可以用标记位数组或者HashSet来表示两个字符串是否重合。
代码
class Solution {
public int maxProduct(String[] words) {
if (words == null || words.length == 0)
return 0;
int len = words.length;
int[] value = new int[len];
for (int i = 0; i < len; i++) {
String tmp = words[i];
value[i] = 0;
for (int j = 0; j < tmp.length(); j++) {
value[i] |= 1 << (tmp.charAt(j) - 'a'); // 这里是把1往左移动 " ch - a" 位,标记0--25位上的1确定a---z是否出现,value数组就是标记数组
}
}
int maxProduct = 0;
for (int i = 0; i < len; i++)
for (int j = i + 1; j < len; j++) {
if ((value[i] & value[j]) == 0
&& (words[i].length() * words[j].length() > maxProduct))
maxProduct = words[i].length() * words[j].length();
}
return maxProduct;
}
}