Merge pull request #597 from Sunjae95/main

[선재] Week14

Author: SamTheKorean

binary-tree-level-order-traversal/sunjae95.js

@@ -0,0 +1,31 @@
+/**
+ * @description
+ * 동일한 depth를 방문해야하므로 bfs 및 트리 순회
+ *
+ * n = length of node of root
+ * time complexity: O(n)
+ * space complexity: O(n)
+ */
+var levelOrder = function (root) {
+  if (!root) return [];
+
+  const answer = [];
+  const queue = [root];
+  let queueCurrentIndex = 0;
+
+  while (queue.length > queueCurrentIndex) {
+    answer.push([]);
+    const answerLastIndex = answer.length - 1;
+    const depthEndIndex = queue.length;
+
+    while (depthEndIndex !== queueCurrentIndex) {
+      const tree = queue[queueCurrentIndex++];
+
+      answer[answerLastIndex].push(tree.val);
+      if (tree.left) queue.push(tree.left);
+      if (tree.right) queue.push(tree.right);
+    }
+  }
+
+  return answer;
+};

house-robber-ii/sunjae95.js

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+/**
+ * @description
+ * 점화식: dp[i] = Math.max(dp[i - 1], dp[i - 2] + current);
+ * 순회한다는 조건이 있으므로 다음과 같이 분기처리 할 수 있다.
+ * 1. 처음이 선택 O 마지막 X
+ * 2. 처음이 선택 X 마지막 상관없음
+ *
+ * n = length of nums
+ * time complexity: O(n)
+ * space complexity: O(n)
+ */
+var rob = function (nums) {
+  if (nums.length === 1) return nums[0];
+  if (nums.length === 2) return Math.max(nums[0], nums[1]);
+
+  const hasFirst = Array.from({ length: nums.length }, (_, i) =>
+    i < 2 ? nums[0] : 0
+  );
+  const noFirst = Array.from({ length: nums.length }, (_, i) =>
+    i === 1 ? nums[i] : 0
+  );
+  for (let i = 2; i < nums.length; i++) {
+    hasFirst[i] = Math.max(hasFirst[i - 1], hasFirst[i - 2] + nums[i]);
+    noFirst[i] = Math.max(noFirst[i - 1], noFirst[i - 2] + nums[i]);
+    if (i === nums.length - 1) {
+      hasFirst[i] = Math.max(hasFirst[i - 1], hasFirst[i - 2]);
+    }
+  }
+
+  return Math.max(hasFirst[nums.length - 1], noFirst[nums.length - 1]);
+};

reverse-bits/sunjae95.js

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+/**
+ * @description
+ *
+ * n = length of n
+ * time complexity: O(n)
+ * space complexity: O(n)
+ */
+var reverseBits = function (n) {
+  let answer = 0;
+  let binary = n.toString(2);
+
+  if (binary.length < 32) binary = "0".repeat(32 - binary.length) + binary;
+
+  for (let i = binary.length - 1; i >= 0; i--)
+    answer += Math.pow(2, i) * Number(binary[i]);
+
+  return answer;
+};