318. 最大单词长度乘积
给你一个字符串数组 words ,找出并返回 length(words[i]) * length(words[j]) 的最大值,并且这两个单词不含有公共字母。如果不存在这样的两个单词,返回 0 。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/maximum-product-of-word-lengths
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位运算
class Solution {
private int toInt(String word) {
int ans = 0;
for (int i = 0; i < word.length(); ++i) {
ans |= (1 << (word.charAt(i) - 'a'));
}
return ans;
}
public int maxProduct(String[] words) {
if (words == null || words.length == 0) {
return 0;
}
int[][] hash = new int[words.length][2];
for (int i = 0; i < words.length; ++i) {
hash[i][0] = words[i].length();
hash[i][1] = toInt(words[i]);
}
int ans = 0;
for (int i = 0; i < hash.length - 1; ++i) {
for (int j = i + 1; j < hash.length; ++j) {
if ((hash[i][1] & hash[j][1]) == 0) {
ans = Math.max(ans, hash[i][0] * hash[j][0]);
}
}
}
return ans;
}
}
位运算优化
import java.util.HashMap;
import java.util.Map;
class Solution {
private int toInt(String word) {
int ans = 0;
for (int i = 0; i < word.length(); ++i) {
ans |= (1 << (word.charAt(i) - 'a'));
}
return ans;
}
public int maxProduct(String[] words) {
if (words == null || words.length == 0) {
return 0;
}
Map<Integer, Integer> maskLengthMap = new HashMap<>();
for (int i = 0; i < words.length; ++i) {
int mask = toInt(words[i]);
maskLengthMap.put(mask, Math.max(words[i].length(), maskLengthMap.getOrDefault(mask, 0)));
}
int ans = 0;
for (Map.Entry<Integer, Integer> maskEntry1 : maskLengthMap.entrySet()) {
for (Map.Entry<Integer, Integer> maskEntry2 : maskLengthMap.entrySet()) {
if ((maskEntry1.getKey() & maskEntry2.getKey()) == 0) {
ans = Math.max(ans, maskEntry1.getValue() * maskEntry2.getValue());
}
}
}
return ans;
}
}
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