11주차 solved

Author: hsskey

binary-tree-maximum-path-sum/hsskey.js

@@ -0,0 +1,36 @@
+/**
+ * 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} root
+ * @return {number}
+ */
+var maxPathSum = function(root) {
+    let res = [root.val];
+
+    function dfs(node) {
+        if (!node) return 0;
+
+        let leftMax = dfs(node.left);
+        let rightMax = dfs(node.right);
+
+        leftMax = Math.max(leftMax, 0);
+        rightMax = Math.max(rightMax, 0);
+
+        // 경유지점 포함한 경로 최대값 업데이트
+        res[0] = Math.max(res[0], node.val + leftMax + rightMax);
+
+        // 분기하지 않는 경로에서의 최대값 리턴
+        return node.val + Math.max(leftMax, rightMax);
+    }
+
+    dfs(root);
+    return res[0];
+};
+

graph-valid-tree/hsskey.js

@@ -0,0 +1,38 @@
+export class Solution {
+    /**
+     * @param {number} n - number of nodes
+     * @param {number[][]} edges - undirected edges
+     * @return {boolean}
+     */
+    validTree(n, edges) {
+      if (n === 0) return true;
+  
+      // 인접 리스트 생성
+      const adj = {};
+      for (let i = 0; i < n; i++) {
+        adj[i] = [];
+      }
+      for (const [n1, n2] of edges) {
+        adj[n1].push(n2);
+        adj[n2].push(n1);
+      }
+  
+      const visit = new Set();
+  
+      const dfs = (i, prev) => {
+        if (visit.has(i)) return false;
+  
+        visit.add(i);
+  
+        for (const j of adj[i]) {
+          if (j === prev) continue;
+          if (!dfs(j, i)) return false;
+        }
+  
+        return true;
+      };
+  
+      return dfs(0, -1) && visit.size === n;
+    }
+  }
+

merge-intervals/hsskey.js

@@ -0,0 +1,26 @@
+/**
+ * @param {number[][]} intervals
+ * @return {number[][]}
+ */
+var merge = function(intervals) {
+    if (intervals.length === 0) return [];
+
+    // 시작값 기준으로 정렬
+    intervals.sort((a, b) => a[0] - b[0]);
+
+    const output = [intervals[0]];
+
+    for (let i = 1; i < intervals.length; i++) {
+        const [start, end] = intervals[i];
+        const lastEnd = output[output.length - 1][1];
+
+        if (start <= lastEnd) {
+            output[output.length - 1][1] = Math.max(lastEnd, end);
+        } else {
+            output.push([start, end]);
+        }
+    }
+
+    return output;
+};
+

missing-number/hsskey.js

@@ -0,0 +1,14 @@
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var missingNumber = function(nums) {
+    let res = nums.length;
+
+    for (let i = 0; i < nums.length; i++) {
+        res += i - nums[i];
+    }
+
+    return res;
+};
+

reorder-list/hsskey.js

@@ -0,0 +1,45 @@
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+
+/**
+ * @param {ListNode} head
+ * @return {void} Do not return anything, modify head in-place instead.
+ */
+var reorderList = function(head) {
+    if (!head || !head.next) return;
+
+    let slow = head, fast = head.next;
+    while (fast && fast.next) {
+        slow = slow.next;
+        fast = fast.next.next;
+    }
+
+    let second = slow.next;
+    let prev = null;
+    slow.next = null;
+    while (second) {
+        let tmp = second.next;
+        second.next = prev;
+        prev = second;
+        second = tmp;
+    }
+
+    let first = head;
+    second = prev;
+    while (second) {
+        let tmp1 = first.next;
+        let tmp2 = second.next;
+
+        first.next = second;
+        second.next = tmp1;
+
+        first = tmp1;
+        second = tmp2;
+    }
+};
+