Merge pull request #1588 from uraflower/main

[uraflower] Week12 Solutions

Author: uraflower

non-overlapping-intervals/uraflower.js

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+/**
+ * 구간이 겹치지 않게 하기 위해 최소한의 구간을 없애야 할 때, 몇 개를 없애야 하는지 반환하는 함수
+ * @param {number[][]} intervals
+ * @return {number}
+ */
+const eraseOverlapIntervals = function(intervals) {
+    intervals.sort((a,b) => a[0] - b[0]);
+    
+    let count = 0;
+    let prevEnd = intervals[0][1];
+
+    for (let i = 1; i < intervals.length; i++) {
+        const [start, end] = intervals[i];
+
+        // 범위가 겹치는 경우
+        if (prevEnd > start) {
+            count++;
+            prevEnd = Math.min(prevEnd, end); // end가 큰 걸 삭제한다 치기
+        }
+    }
+    
+    return count;
+};
+
+// 시간복잡도: O(n * log n)
+// 공간복잡도: O(1)

number-of-connected-components-in-an-undirected-graph/uraflower.js

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+
+/**
+ * @param n: the number of vertices
+ * @param edges: the edges of undirected graph
+ * @return: the number of connected components
+ */
+const countComponents = function (n, edges) {
+  // graph 만들기
+  const graph = Array.from({ length: n }).map(() => []);
+
+  for (const [u, v] of edges) {
+    graph[u].push(v);
+    graph[v].push(u);
+  }
+
+  // 각 노드 순회하기
+  let count = 0;
+  const visited = new Set();
+
+  for (let i = 0; i < n; i++) {
+    if (visited.has(i)) {
+      continue;
+    }
+
+    count += 1;
+
+    // bfs
+    const queue = [i];
+    visited.add(i);
+
+    while (queue.length) {
+      const u = queue.shift();
+
+      for (const v of graph[u]) {
+        if (!visited.has(v)) {
+          visited.add(v);
+          queue.push(v);
+        }
+      }
+    }
+  }
+
+  return count;
+}
+
+// 시간복잡도: O(E)
+// 공간복잡도: O(V)

remove-nth-node-from-end-of-list/uraflower.js

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+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+
+// 첫 번째 풀이
+// 시간복잡도: O(n)
+// 공간복잡도: O(n)
+/**
+ * @param {ListNode} head
+ * @param {number} n
+ * @return {ListNode}
+ */
+const removeNthFromEnd = function (head, n) {
+    function dfs(node) {
+        // 마지막 노드이면 1 반환
+        if (!node.next) {
+            return 1;
+        }
+
+        const nth = dfs(node.next);
+        if (nth === n) {
+            node.next = node.next.next ?? null;
+        }
+
+        return nth + 1;
+    }
+
+    if (dfs(head) === n) {
+        return head.next;
+    }
+
+    return head;
+};
+
+// 두 번째 풀이
+// 시간복잡도: O(n)
+// 공간복잡도: O(1)
+const removeNthFromEnd = function (head, n) {
+    const temp = new ListNode(0, head);
+    let tail = temp; 
+    let prev = temp; // 뒤에서 n번째 노드
+    
+    for (let i = 0; i < n; i++) {
+        tail = tail.next;
+    }
+
+    while (tail.next) {
+        tail = tail.next;
+        prev = prev.next;
+    }
+    
+    prev.next = prev.next.next;
+
+    return temp.next;
+};

same-tree/uraflower.js

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+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} p
+ * @param {TreeNode} q
+ * @return {boolean}
+ */
+const isSameTree = function(p, q) {
+    if (!p && !q) return true;
+    return p?.val === q?.val && isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
+};
+
+// 시간복잡도: O(n)
+// 공간복잡도: O(h) (h: 트리의 높이, 즉 재귀 호출 스택)

serialize-and-deserialize-binary-tree/uraflower.js

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+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val) {
+ *     this.val = val;
+ *     this.left = this.right = null;
+ * }
+ */
+
+const NULL_SIGN = 'X';
+
+/**
+ * Encodes a tree to a single string.
+ * 시간복잡도: O(n)
+ * 공간복잡도: O(h) (h: 재귀 스택 깊이 즉 트리 높이)
+ * @param {TreeNode} root
+ * @return {string}
+ */
+const serialize = function (root) {
+    const result = [];
+
+    function traverse(root) {
+        result.push(root?.val ?? NULL_SIGN);
+        if (!root) {
+            return;
+        }
+        traverse(root.left);
+        traverse(root.right);
+    }
+
+    traverse(root);
+    return result.join(',');
+};
+
+/**
+ * Decodes your encoded data to tree.
+ * 시간복잡도: O(n)
+ * 공간복잡도: O(h) (h: 재귀 스택 깊이 즉 트리 높이)
+ * @param {string} data
+ * @return {TreeNode}
+ */
+const deserialize = function (data) {
+    const splited = data.split(',');
+    let i = 0;
+
+    function makeTree() {
+        if (splited[i] === NULL_SIGN) {
+            return null;
+        }
+
+        const node = new TreeNode(Number(splited[i]));
+        i += 1;
+        node.left = makeTree();
+        i += 1;
+        node.right = makeTree();
+
+        return node;
+    }
+
+    return makeTree();
+};
+
+/**
+ * Your functions will be called as such:
+ * deserialize(serialize(root));
+ */