Maximum Product of Word Lengths
Total Accepted: 16297 Total Submissions: 41606 Difficulty: Medium
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:
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.
思路
- brute force 直接强行比较两个字符串看是否有相同char,会比较慢
- 所以想到了,先用hashmap来存好每个字符串sort过后的char[] array,然后用两根指针来找是否有相同char
- 时间O(n2) 空间O(n)
- 不过还不是最好的解法,我们可以使用bit map来解决这道题
public class Solution {
public int maxProduct(String[] words) {
if (words == null || words.length == 0) {
return 0;
}
Map<String, char[]> map = new HashMap<String, char[]>();
int n = words.length;
int result = 0;
for (int i = 0; i < n; i++) {
String cur = words[i];
char[] sortArray = cur.toCharArray();
Arrays.sort(sortArray);
map.put(cur, sortArray);
}
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
String s1 = words[i];
String s2 = words[j];
if (shareSameChar(map.get(s1), map.get(s2))) {
continue;
}
result = Math.max(result, s1.length() * s2.length());
}
}
return result;
}
public boolean shareSameChar(char[] array1, char[] array2) {
int start1 = 0;
int start2 = 0;
while (start1 < array1.length && start2 < array2.length) {
if (array1[start1] == array2[start2]) {
return true;
} else if (array1[start1] > array2[start2]) {
start2++;
} else {
start1++;
}
}
return false;
}
}
优化
- 因为题目中说都是小写字母,那么只有26位,一个整型数int有32位,我们可以用后26位来对应26个字母,若为1,说明该对应位置的字母出现过,那么每个单词的都可由一个int数字表示,两个单词没有共同字母的条件是这两个int数想与为0
public class Solution {
public int maxProduct(String[] words) {
int n = words.length;
int[] elements = new int[n];
for (int i=0;i<n;i++){
for(int j=0;j<words[i].length();j++){
elements[i] |= 1 << (words[i].charAt(j) - 'a');
}
}
int ans = 0;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if ((elements[i] & elements[j]) == 0)
ans = Math.max(ans,words[i].length() * words[j].length());
}
}
return ans;
}
}