[LeetCode]Maximum Product of Word Lengths

题目描述

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.

代码如下：

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public class Solution { public int (String[] words) { if(words == null || words.length == 0) return 0; int length = words.length; int[] masks = new int[length]; //使用masks[i]记录words[i]含有的字符信息 for(int i = 0; i < length; i++){ //遍历words[i]的每个字符，masks[i]的最低比特位表示words[i]是否含有字符a(1表示含有 //，0表示不含)，最高位表示是否含有字符z for(char ch: words[i].toCharArray()){ masks[i] |= 1 << (ch - 'a'); } } int maxProduct = 0; for(int i = 0; i < length; i++){ for(int j = i+1; j < length; j++){ //(masks[i] & masks[j]) == 0表示两个word没有相同的字符 if((masks[i] & masks[j]) == 0){ maxProduct = Math.max(maxProduct, words[i].length()*words[j].length()); } } } return maxProduct; } }

或者优化一下：

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//我们还可以先对words数组按字符串长度排序，后面就可以进行剪枝优化,排序为O(nlgn)，不影响O(n*n)的时间复杂度 public int (String[] words) { if(words == null || words.length == 0) return 0; //对words数组按字符串长度排序 Arrays.sort(words, new Comparator<String>(){ public int compare(String a, String b){ return b.length() - a.length(); } }); int length = words.length; int[] masks = new int[length]; //使用masks[i]记录words[i]含有的字符信息 for(int i = 0; i < length; i++){ //遍历words[i]的每个字符，masks[i]的最低比特位表示words[i]是否含有字符a(1表示含有 //，0表示不含)，最高位表示是否含有字符z for(char ch: words[i].toCharArray()){ masks[i] |= 1 << (ch - 'a'); } } int maxProduct = 0; for(int i = 0; i < length; i++){ //words[i].length() * words[i].length()是words[i]最大可能的maxProduct if(words[i].length() * words[i].length() < maxProduct) break; //剪枝优化 for(int j = i+1; j < length; j++){ //(masks[i] & masks[j]) == 0表示两个word没有相同的字符 if((masks[i] & masks[j]) == 0){ maxProduct = Math.max(maxProduct, words[i].length()*words[j].length()); break;//剪枝，越前面的字符串求到的maxProduct越大 } } } return maxProduct; }