Merge pull request #1707 from yg-cho/main

[Yg-cho] WEEK 01 Solutions

Author: yg-cho

contains-duplicate/Yg-cho.js

@@ -0,0 +1,11 @@
+/**
+ * @param {number[]} nums
+ * @return {boolean}
+ */
+var containsDuplicate = function(nums) {
+  return new Set(nums).size !== nums.length;
+};
+
+//console.log(containsDuplicate([1, 2, 3, 1])); // true
+// console.log(containsDuplicate([1, 2, 3, 4])); // false
+// console.log(containsDuplicate([1, 1, 1, 3, 3, 4, 3, 2, 4, 2])); // true

longest-consecutive-sequence/Yg-cho.js

@@ -0,0 +1,25 @@
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var longestConsecutive = function(nums) {
+  if(nums.length === 0) return 0;
+
+  const numSet = new Set(nums);
+  let longest = 0;
+
+  for (const num of numSet) {
+    if(!numSet.has(num-1)) {
+      let currentNum = num;
+      let currentLength = 1;
+
+      while(numSet.has(currentNum+1)) {
+        currentNum++;
+        currentLength++;
+      }
+
+      longest = Math.max(longest, currentLength)
+    }
+  }
+  return longest;
+};

top-k-frequent-elements/Yg-cho.js

@@ -0,0 +1,19 @@
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number[]}
+ */
+var topKFrequent = function(nums, k) {
+  //HashMap 선언
+  const counter = new Map()
+
+  //HashMap에 빈도를 value로 저장
+  for(const num of nums){
+    counter.set(num,(counter.get(num)|| 0) +1);
+  }
+
+  //keys를 가져와 정렬 후, k만큼 -slice 리턴
+  return [...counter.keys()]
+    .sort((a,b) => counter.get(a) - counter.get(b))
+    .slice(-k)
+};

two-sum/Yg-cho.js

@@ -0,0 +1,19 @@
+/**
+ * @param {number[]} nums
+ * @param {number} target
+ * @return {number[]}
+ */
+var twoSum = function(nums, target) {
+  for(var i = 0; i < nums.length; i++){
+    let getNum = target - nums[i];
+    let foundIndex = nums.indexOf(getNum);
+
+    // foundIndex가 존재하고(-1이 아니고), 자기 자신이 아닌 경우
+    if(foundIndex !== -1 && foundIndex !== i) {
+      return [i, foundIndex];
+    }
+  }
+};
+console.log(twoSum([2, 7, 11, 15], 9)); // [0, 1]
+console.log(twoSum([3, 2, 4], 6));      // [1, 2]
+console.log(twoSum([3, 3], 6));         // [0, 1]