Maximum Product of Word Lengths
題意：
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.
解題思路：
updated 2016.1.9
參考了網友 cc9208 的解答，他用了bitmap來快速幫助我們解決，加上根據字串長度排序了陣列，加上使用pruning，只要兩個字串長度相乘不超過當前的最長長度，則不繼續往下算。
public class Solution {
public int maxProduct(String[] words) {
int max = 0;
Arrays.sort(words, new Comparator<String>(){
public int compare(String a, String b){
return b.length() - a.length();
}
});
int[] masks = new int[words.length]; // alphabet masks
for(int i = 0; i < masks.length; i++){
for(char c: words[i].toCharArray()){
masks[i] |= 1 << (c - 'a');
}
}
for(int i = 0; i < masks.length; i++){
if(words[i].length() * words[i].length() <= max) break; //prunning
for(int j = i + 1; j < masks.length; j++){
if((masks[i] & masks[j]) == 0){
max = Math.max(max, words[i].length() * words[j].length());
break; //prunning
}
}
}
return max;
}
}
不斷的拿一個字與後面的所有字來作比較，用一個陣列來紀錄當前的字有哪些字母，一但遇到有相同的字母則直接跳出，否則計算出兩個字串的長度乘積是否與目前max還大。
public class Solution {
public int maxProduct(String[] words) {
if (words == null || words.length == 0) {
return 0;
}
int res = 0;
for (int i = 0; i < words.length; i++) {
int[] letters = new int[26];
for (char c : words[i].toCharArray()) {
letters[c - 'a']++;
}
for (int j = i + 1; j < words.length; j++) {
int k = 0;
for (;k < words[j].length(); k++) {
if (letters[words[j].charAt(k) - 'a'] != 0) {
break;
}
}
if (k == words[j].length()) {
res = Math.max(res, words[i].length() * words[j].length());
}
}
}
return res;
}
}