Merge pull request #1239 from froggy1014/main

[froggy1014] WEEK 02 solutions

Author: froggy1014

climbing-stairs/froggy1014.js

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+/**
+ * @param {number} n
+ * @return {number}
+ */
+var climbStairs = function(n) {
+    if (n === 1) return 1;
+    if (n === 2) return 2;
+    
+    let dp = new Array(n + 1).fill(0);
+    dp[1] = 1;
+    dp[2] = 2;
+    for (let i = 3; i <= n; i++) {
+        dp[i] = dp[i - 1] + dp[i - 2];
+    }
+    return dp[n];
+};
+
+
+console.log(climbStairs(4));

product-of-array-except-self/froggy1014.js

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+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var productExceptSelf = function (nums) {
+  const array = new Array(nums.length).fill(1);
+
+  for (let i = 1; i < nums.length; i++) {
+    array[i] = array[i - 1] * nums[i - 1];
+  }
+
+  let right = 1;
+  for (let i = nums.length - 1; i >= 0; i--) {
+    array[i] = array[i] * right;
+    right *= nums[i];
+  }
+
+  return array;
+};
+
+console.log(productExceptSelf([1, 2, 3, 4]));

valid-anagram/froggy1014.js

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+
+// 맵 자료 구조를 통해서 각 문자의 개수를 카운트 해서 비교
+var isAnagram = function (s, t) {
+  const map = new Map();
+
+  for (let i = 0; i < s.length; i++) {
+    map.set(s[i], (map.get(s[i]) || 0) + 1);
+  }
+
+  for (let i = 0; i < t.length; i++) {
+    map.set(t[i], (map.get(t[i]) || 0) - 1);
+  }
+
+  return Array.from(map.values()).every((value) => value === 0);
+};
+
+console.log(isAnagram("anagram", "nagaram"));
+console.log(isAnagram("rat", "car"));