Merge pull request #157 from clingoram/mavis

新增解題

Author: clingoram

javascript/LeetCode/Array/1967.js

@@ -0,0 +1,31 @@
+/**
+ * 1967. Number of Strings That Appear as Substrings in Word
+ * 
+ * @param {string[]} patterns
+ * @param {string} word
+ * @return {number}
+ */
+var numOfStrings = function(patterns, word) {
+    /**
+     * 計算patterns中有幾個元素出現在word中
+     */
+    let count = 0;
+    for(let i = 0;i < patterns.length;++i) {
+      if(word.includes(patterns[i])){
+        count++;
+      }
+    }
+    return count;
+};
+let patterns = ["a","abc","bc","d"], word = "abc";
+/*
+Output: 3
+Explanation:
+- "a" appears as a substring in "abc".
+- "abc" appears as a substring in "abc".
+- "bc" appears as a substring in "abc".
+- "d" does not appear as a substring in "abc".
+3 of the strings in patterns appear as a substring in word.
+*/
+console.log(numOfStrings(patterns,word));
+

javascript/LeetCode/Array/804.js

@@ -0,0 +1,34 @@
+/**
+ * 804. Unique Morse Code Words
+ * 
+ * 給一個字串陣列作為參數，把該陣列內的元素一一拆開成字元，去比對每個字元轉換成摩斯密碼後共有幾個不同的
+ * 
+ * 英文字母是小寫
+ * @param {string[]} words
+ * @return {number}
+ */
+var uniqueMorseRepresentations = function(words) {
+  // 26英文字母分別代表的morse code
+  let morse = [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."];
+
+  let set = new Set();
+  for(const letter of words){
+    let combine = "";
+    for(const a of letter) {
+      combine += morse[a.charCodeAt(0) - 'a'.charCodeAt(0)];
+    }
+    set.add(combine);
+  }
+  return set.size;
+};
+let words = ["gin","zen","gig","msg"]
+/*
+Output: 2
+Explanation: The transformation of each word is:
+"gin" -> "--...-."
+"zen" -> "--...-."
+"gig" -> "--...--."
+"msg" -> "--...--."
+There are 2 different transformations: "--...-." and "--...--.".
+*/
+console.log(uniqueMorseRepresentations(words));

javascript/LeetCode/String/3340.js

@@ -0,0 +1,23 @@
+/**
+ * 3340. Check Balanced String
+ * 
+ * 參數為字串數字，檢查index 偶數和基數各自相加後是否相等。
+ * 
+ * @param {string} num
+ * @return {boolean}
+ */
+var isBalanced = function(num) {
+    // solution 
+    // TC:O(N)
+    let oddSum = 0, evenSum = 0;
+    for(let i = 0;i < num.length;++i) {
+        if(i % 2 === 0){
+            evenSum += parseInt(num[i]);
+        }else{
+            oddSum += parseInt(num[i]);
+        }
+    }
+    return evenSum === oddSum;
+};
+let num = "1234";
+console.log(isBalanced(num));

javascript/index.js

@@ -836,7 +836,7 @@ var longestPalindrome = function(words) {
 
 
 };
-let words = ["ab","ty","yt","lc","cl","ab"];
+// let words = ["ab","ty","yt","lc","cl","ab"];
 // 8
 // "ty" + "lc" + "cl" + "yt" = "tylc clyt"
 // abtyytba
@@ -1383,4 +1383,50 @@ Query 0: The element at queries[0] = 0 is nums[0] = 1. The nearest index with th
 Query 1: The element at queries[1] = 3 is nums[3] = 4. No other index contains 4, so the result is -1.
 Query 2: The element at queries[2] = 5 is nums[5] = 3. The nearest index with the same value is 1, and the distance between them is 3 (following the circular path: 5 -> 6 -> 0 -> 1).
 */
-// console.log(solveQueries(nums,queries));
+// console.log(solveQueries(nums,queries));
+
+/**
+ * 645. Set Mismatch
+ * 
+ * 參數為數字陣列，從1至n，但內有重複的元素，找出它們並調整成對的元素
+ *
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var findErrorNums = function(nums) {
+  let res = [];
+  let map = new Map();
+  for(const ele of nums){
+    map.has(ele) ? map.set(ele,map.get(ele) + 1 ) : map.set(ele,1);
+  }
+  // console.log(map)
+  for(const [key,value] of map) {
+    if(value >= 2){
+      console.log(key);
+
+    }
+  }
+};
+let nums = [1,2,2,4];
+// [2,3]
+// console.log(findErrorNums(nums));
+
+
+/**
+ * 2864. Maximum Odd Binary Number
+ * 
+ * @param {string} s
+ * @return {string}
+ */
+var maximumOddBinaryNumber = function(s) {
+    /**
+    * binary string s包含至少一個'1'
+    * rearrange 參數s成最大的奇二進制數
+    * 回傳的結果值開頭可以是0
+    */
+   
+};
+let s = "010";
+// Output: "001"
+// Explanation: Because there is just one '1', it must be in the last position. So the answer is "001".
+console.log(maximumOddBinaryNumber(s));