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[환미니니] 8주차 문제풀이 제출 #509

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Oct 5, 2024
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39 changes: 39 additions & 0 deletions clone-graph/hwanmini.js
Original file line number Diff line number Diff line change
@@ -0,0 +1,39 @@
// 노드 수 v, 간선 e
// 시간복잡도 O(v+e)
// 공간복잡도 O(v+e)

/**
* // Definition for a _Node.
* function _Node(val, neighbors) {
* this.val = val === undefined ? 0 : val;
* this.neighbors = neighbors === undefined ? [] : neighbors;
* };
*/

/**
* @param {_Node} node
* @return {_Node}
*/
var cloneGraph = function(node) {
if (!node) return

const visited = new Map()
visited.set(node, new _Node(node.val))

const que = [node]
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qqueue로 표현해준다면 더 가독성이 높아질 것 같아요!

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변수명 좀 더 신경써서 지어야겠어요👍


while (que.length) {
const curNode = que.shift()

for (neighbor of curNode.neighbors) {
if (!visited.has(neighbor)) {
visited.set(neighbor, new _Node(neighbor.val));
que.push(neighbor)
}

visited.get(curNode).neighbors.push(visited.get(neighbor))
}
}
Comment on lines +25 to +36
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👍


return visited.get(node)
};
37 changes: 37 additions & 0 deletions merge-two-sorted-lists/hwanmini.js
Original file line number Diff line number Diff line change
@@ -0,0 +1,37 @@
// 시간복잡도: O(m + n)
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clone graph 문제에서 해주신 것처럼 m, n을 무엇으로 정의했는지 적어주면 더 좋을 것 같아요.

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좋습니다 ~

// 공간복잡도: O(m + n)

/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} list1
* @param {ListNode} list2
* @return {ListNode}
*/
var mergeTwoLists = function(list1, list2) {
let res = new ListNode()
let resCopy = res

while (list1 && list2) {
if (list1.val < list2.val) {
res.next = new ListNode(list1.val);
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지금은 순회를 할때마다 새로운 노드를 만들어서 할당하고 있어요.
이걸 개선할 수 있는 방법이 있을까요~? 🤔

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아 기존 리스트 노드를 재활용하면 되겠네요..!
감사합니다 : )

list1 = list1.next;
} else {
res.next = new ListNode(list2.val);
list2 = list2.next;
}

res = res.next
}

if (list1) res.next = list1;
if (list2) res.next = list2

return resCopy.next
};