Example 1:
Given ["abcw", "baz", "foo", "bar", "xtfn", "abcdef"]
Return 16
The two words can be "abcw", "xtfn".
Example 2:
Given ["a", "ab", "abc", "d", "cd", "bcd", "abcd"]
Return 4
The two words can be "ab", "cd".
Example 3:
Given ["a", "aa", "aaa", "aaaa"]
Return 0
No such pair of words.
给定一个字符串数组,从中找出两个字符串使它们的乘积最大,但是这两个字符串中不存在相同的字符。如果直接查找,数组中每两个字符串都要相互比较,字符串中的字符也要比较,时间复杂度太大。我们可以采用位运算,先将字符串中出现的字符用二进制位来表示,一共有26个字符,我们用一个32位的整数就可以表示。然后判断两个字符串中是否存在相同的字符时,可以将两个字符串的对应的二进制数进行位与运算,如果为0说明不存在相同字符,如果不为0说明存在相同字符。时间复杂度为O(n^2)。代码如下:
public class Solution { public int maxProduct(String[] words) { if(words == null || words.length == 0) return 0; int max = 0; int[] bitString = new int[words.length]; for(int i = 0; i < words.length; i++) { char[] w = words[i].toCharArray(); for(int j = 0; j < w.length; j++) { bitString[i] |= 1 << (w[j] - 'a'); } } for(int i = 0; i < bitString.length - 1; i++) { for(int j = i + 1; j < bitString.length; j++) { if((bitString[i] & bitString[j]) == 0) max = Math.max(max, words[i].length() * words[j].length()); } } return max; } }