feixiangcode
diff --git a/‎Week_02/id_130/LeetCode_242_130.js‎
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diff --git a/‎Week_02/id_130/LeetCode_692_130.js‎
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diff --git a/‎Week_02/id_130/NOTE.md‎
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@@ -0,0 +1,24 @@
+// https://leetcode.com/problems/valid-anagram/
+
+/**
+ * @param {string} s
+ * @param {string} t
+ * @return {boolean}
+ */
+var isAnagram = function(s, t) {
+    if (s.length !== t.length) return false;
+    const countS = new Array(26).fill(0);
+    const countT = new Array(26).fill(0);
+    for (let i = 0; i < s.length; i++) {
+        countS[s.charCodeAt(i) - 97]++;
+        countT[t.charCodeAt(i) - 97]++;
+    }
+    
+    for (let i = 0; i < 26; i++) {
+        if (countS[i] !== countT[i]) {
+            return false;
+        }
+    }
+    
+    return true;
+};
@@ -0,0 +1,29 @@
+// https://leetcode.com/problems/top-k-frequent-words/
+
+/**
+ * @param {string[]} words
+ * @param {number} k
+ * @return {string[]}
+ */
+var topKFrequent = function(words, k) {
+    const counts = {};
+    for(let i = 0; i < words.length; i++) {
+        counts[words[i]] ? counts[words[i]]++ : counts[words[i]] = 1;
+    }
+    
+     const keys = Object.keys(counts).sort((a, b) => {
+        if (counts[a] === counts[b]) {
+            if (a > b) {
+                return 1;
+            } else {
+                return -1;
+            }
+        }
+        else {
+            return counts[b] - counts[a];
+        }
+	})
+	.slice(0, k);   
+    
+    return keys.slice(0, k);
+};
@@ -1 +1,13 @@
-# 学习笔记
+# 学习笔记
+
+#### 散列表
+
+- 查询时间复杂度 O(1)，所以主要用于计数，需要配合排序及其他算法使用
+
+#### 跳表
+
+- 利用多级索引，提高查找效率
+
+#### 二叉树
+
+- 递归算法