LeetCode（Map)804. Unique Morse Code Words
问题
International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows:
‘a’ maps to “.-”, ‘b’ maps to “-…”, ‘c’ maps to “-.-.”, and so on. For convenience, the full table for the 26 letters of the English alphabet is given below:
[“.-”,“-…”,“-.-.”,“-…”,“.”,“…-.”,“–.”,“…”,“…”,“.—”,“-.-”,“.-…”,“–”,“-.”,“—”,“.–.”,“–.-”,“.-.”,“…”,“-”,“…-”,“…-”,“.–”,“-…-”,“-.–”,“–…”] Given an array of strings words where each word can be written as a concatenation of the Morse code of each letter.
For example, “cab” can be written as “-.-…–…”, which is the concatenation of “-.-.”, “.-”, and “-…”. We will call such a concatenation the transformation of a word. Return the number of different transformations among all words we have.
Example 1:
Input: words = [“gin”,“zen”,“gig”,“msg”]
Output: 2
Explanation: The transformation of each word is:
“gin” -> “–…-.”
“zen” -> “–…-.”
“gig” -> “–…–.”
“msg” -> “–…–.”
There are 2 different transformations: “–…-.” and “–…–.”.
Example 2:
Input: words = [“a”]
Output: 1
Constraints:
1 <= words.length <= 100 1 <= words[i].length <= 12
代码
class Solution {
public int uniqueMorseRepresentations(String[] words) {
String[] tokens = new String[]{".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
Set
for ( String word : words ) {//2.遍历数组
StringBuilder sb = new StringBuilder();//使用 StringBuilder是可以添加不限长度的元素
for ( int i = 0; i < word.length(); i++ ) {//ASC2码小写字母位置是连续的，所以 ‘字母’-’a’=字母和a的距离，此题就对应morse数组中对应字母的索引,并添加到StringBUilder
sb.append( tokens[word.charAt(i)-'a'] );//int index=c-'a';sb.append(morseCode[index]);
}
set.add( sb.toString() );//转换成String添加到hashset
}
return set.size();//3.求hashset的size也就是不重复的元素个数
}
}
public int uniqueMorseRepresentations(String[] words) {
String[] arr = new String[]{".-", "-...", "-.-.", "-..", ".", "..-.", "--.", "....", "..", ".---", "-.-", ".-..", "--", "-.", "---", ".--.", "--.-", ".-.", "...", "-", "..-", "...-", ".--", "-..-", "-.--", "--.."};
Set
for (String s : words) {
StringBuilder sb = new StringBuilder();
for (char ch : s.toCharArray()) sb.append(arr[ch - 'a']);//ASC2码小写字母位置是连续的，所以 ‘字母’-’a’=字母和a的距离，此题就对应morse数组中对应字母的索引
set.add(sb.toString());
}
return set.size();
}
LeetCode（Map)2006. Count Number of Pairs With Absolute Difference K
问题
Given an integer array nums and an integer k, return the number of pairs (i, j) where i < j such that |nums[i] - nums[j]| == k.
The value of |x| is defined as:
x if x >= 0. -x if x < 0.
Example 1:
Input: nums = [1,2,2,1], k = 1
Output: 4
Explanation: The pairs with an absolute difference of 1 are:
[1,2,2,1]
[1,2,2,1]
[1,2,2,1]
[1,2,2,1]
Example 2:
Input: nums = [1,3], k = 3
Output: 0
Explanation: There are no pairs with an absolute difference of 3.
Example 3:
Input: nums = [3,2,1,5,4], k = 2
Output: 3
Explanation: The pairs with an absolute difference of 2 are:
[3,2,1,5,4]
[3,2,1,5,4]
[3,2,1,5,4]
Constraints:
1 <= nums.length <= 200 1 <= nums[i] <= 100 1 <= k <= 99
代码
class Solution {
public int countKDifference(int[] nums, int k) {
int result =0;//1.定义result初始为0
for (int i = 0; i < nums.length; i++) {//2.for循环，两数相减取绝对值，如果等于k,result基础上加1
for(int j=i+1; j if(Math.abs(nums[i]-nums[j])==k) result+=1; // if(nums[i]-nums[j]==k) result+=1; // else if( nums[i]-nums[j]==-k) result+=1; } } return result;//3.返回result结果 } } class Solution { public int countKDifference(int[] nums, int k) { Map int result = 0; for(int i = 0;i< nums.length;i++){//2.for循环遍历 if(map.containsKey(nums[i]-k)){//map中是否有key减去k，说明有和其相减的数值，如果有，map.get(nums[i]-k)为1，也就是在result基础上加1 result += map.get(nums[i]-k); } if(map.containsKey(nums[i]+k)){//map中是否有key加k，说明有和其相减的数值，如果有，map.get(nums[i]+k)为1，也就是在result基础上加1 result += map.get(nums[i]+k); } map.put(nums[i],map.getOrDefault(nums[i],0)+1);//map中没有元素，map将数组元素作为key放入map,默认值为0，+1是为reusult加1做铺垫 } return result;//3.返回reuslt } } LeetCode（Map)1684. Count the Number of Consistent Strings 问题 You are given a string allowed consisting of distinct characters and an array of strings words. A string is consistent if all characters in the string appear in the string allowed. Return the number of consistent strings in the array words. Example 1: Input: allowed = “ab”, words = [“ad”,“bd”,“aaab”,“baa”,“badab”] Output: 2 Explanation: Strings “aaab” and “baa” are consistent since they only contain characters ‘a’ and ‘b’. Example 2: Input: allowed = “abc”, words = [“a”,“b”,“c”,“ab”,“ac”,“bc”,“abc”] Output: 7 Explanation: All strings are consistent. Example 3: Input: allowed = “cad”, words = [“cc”,“acd”,“b”,“ba”,“bac”,“bad”,“ac”,“d”] Output: 4 Explanation: Strings “cc”, “acd”, “ac”, and “d” are consistent. Constraints: 1 <= words.length <= 104 1 <= allowed.length <= 26 1 <= words[i].length <= 10 The characters in allowed are distinct. words[i] and allowed contain only lowercase English letters. 代码 class Solution { public int countConsistentStrings(String allowed, String[] words) { int result =0;//1.定义result初始为0 Set for(char a: allowed.toCharArray()){ set.add(a); } for(String w:words){//3.for循环遍历words，如果set包含words中char数组元素为ture，result+1，否者为false，result不变 boolean flag = true; for(char c: w.toCharArray()){ if(!set.contains(c)){//注意：这是set.contains(c)是charter.contains(charter) flag=false; } } if(flag) result+=1; } return result;//4.返回result } } LeetCode（Map)1832. Check if the Sentence Is Pangram 问题 A pangram is a sentence where every letter of the English alphabet appears at least once. Given a string sentence containing only lowercase English letters, return true if sentence is a pangram, or false otherwise. Example 1: Input: sentence = “thequickbrownfoxjumpsoverthelazydog” Output: true Explanation: sentence contains at least one of every letter of the English alphabet. Example 2: Input: sentence = “leetcode” Output: false Constraints: 1 <= sentence.length <= 1000 sentence consists of lowercase English letters. 代码 class Solution { public boolean checkIfPangram(String sentence) { Set set = new HashSet<>();//1.利用set内没有重复元素 for(char c : sentence.toCharArray()) set.add(c);//2.将sentence存入set，如果长度等于26为true，否则为false return set.size() == 26; } } LeetCode（Map)2367. Number of Arithmetic Triplets 问题 You are given a 0-indexed, strictly increasing integer array nums and a positive integer diff. A triplet (i, j, k) is an arithmetic triplet if the following conditions are met: i < j < k, nums[j] - nums[i] == diff, and nums[k] - nums[j] == diff. Return the number of unique arithmetic triplets. Example 1: Input: nums = [0,1,4,6,7,10], diff = 3 Output: 2 Explanation: (1, 2, 4) is an arithmetic triplet because both 7 - 4 == 3 and 4 - 1 == 3. (2, 4, 5) is an arithmetic triplet because both 10 - 7 == 3 and 7 - 4 == 3. Example 2: Input: nums = [4,5,6,7,8,9], diff = 2 Output: 2 Explanation: (0, 2, 4) is an arithmetic triplet because both 8 - 6 == 2 and 6 - 4 == 2. (1, 3, 5) is an arithmetic triplet because both 9 - 7 == 2 and 7 - 5 == 2. Constraints: 3 <= nums.length <= 200 0 <= nums[i] <= 200 1 <= diff <= 50 nums is strictly increasing. 代码 class Solution { public int arithmeticTriplets(int[] nums, int diff) { int result = 0; Set for(int n:nums){ set.add(n); } for(int s:nums){//2.for循环遍历，set是否包含nums数组元素，元素+diff,元素+2*diff，如果有result加1 if(set.contains(s+diff)&&set.contains(s+2*diff)){ result+=1; } } return result;//3.返回result } } 相关链接
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