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Merge pull request #1590 from byol-han/main
[byol-han] WEEK 12 solutions
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/**
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* https://leetcode.com/problems/non-overlapping-intervals/description/
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* @param {number[][]} intervals
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* @return {number}
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*/
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var eraseOverlapIntervals = function (intervals) {
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if (intervals.length === 0) return 0;
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// 1. intervals를 end 값을 기준으로 오름차순 정렬
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intervals.sort((a, b) => a[1] - b[1]);
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// 2. 첫 번째 interval의 end 값으로 초기화
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let end = intervals[0][1];
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let count = 0; // 제거해야 하는 interval의 개수
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// 3. 두 번째 interval부터 순회
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for (let i = 1; i < intervals.length; i++) {
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let [start_i, end_i] = intervals[i];
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if (start_i < end) {
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// 현재 interval의 start가 이전 end보다 작으면 overlap
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count++; // 이 interval은 제거해야 함
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} else {
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// overlap이 없으면 end 갱신
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end = end_i;
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}
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}
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return count;
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};
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/**
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* https://leetcode.com/problems/remove-nth-node-from-end-of-list/submissions/1672151418/
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* Definition for singly-linked list.
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* function ListNode(val, next) {
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* this.val = (val===undefined ? 0 : val)
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* this.next = (next===undefined ? null : next)
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* }
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*/
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/**
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* @param {ListNode} head
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* @param {number} n
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* @return {ListNode}
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*/
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var removeNthFromEnd = function (head, n) {
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// dummy 노드를 만들어 head 앞에 두기 (edge case: 첫 번째 노드를 제거할 때 편리)
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let dummy = new ListNode(0);
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dummy.next = head;
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// fast와 slow 포인터를 dummy에서 시작
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let fast = dummy;
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let slow = dummy;
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// fast 포인터를 n+1 만큼 앞으로 이동시킴
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// (slow가 제거 노드의 바로 이전 노드에 멈추도록 하기 위함)
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for (let i = 0; i < n + 1; i++) {
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fast = fast.next;
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}
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// fast와 slow를 함께 끝까지 이동
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while (fast !== null) {
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fast = fast.next;
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slow = slow.next;
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}
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// slow.next가 제거할 노드이므로, 그 노드를 skip
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slow.next = slow.next.next;
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// dummy.next가 새 head
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return dummy.next;
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};

same-tree/byol-han.js

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/**
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* WEEK12
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* https://leetcode.com/problems/same-tree/description/
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* Definition for a binary tree node.
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* function TreeNode(val, left, right) {

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