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[환미니니] Week4 문제풀이 #425
[환미니니] Week4 문제풀이 #425
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Original file line number | Diff line number | Diff line change |
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// 시간 복잡도 O(n log n) | ||
// 공간 복잡도 O(n) | ||
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/** | ||
* @param {number[]} nums | ||
* @return {number} | ||
*/ | ||
var longestConsecutive = function(nums) { | ||
if (nums.length === 0) return [] | ||
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let maxSequenceLength = -Infinity | ||
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const setNums = [...new Set(nums)].toSorted((a,b) => a - b) | ||
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let count = 0; | ||
for (let i = 0 ; i < setNums.length; i++) { | ||
if (setNums[i]+1 === setNums[i+1]) { | ||
count += 1 | ||
} else { | ||
count += 1 | ||
maxSequenceLength = Math.max(maxSequenceLength, count) | ||
count = 0; | ||
} | ||
} | ||
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return maxSequenceLength | ||
}; | ||
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console.log(longestConsecutive([100,4,200,1,3,2])) | ||
console.log(longestConsecutive([0,3,7,2,5,8,4,6,0,1])) |
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// 시간복잡도 O(n log n) | ||
// 공간복잡도 O(n) | ||
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/** | ||
* @param {number[]} nums | ||
* @return {number} | ||
*/ | ||
var missingNumber = function(nums) { | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. js에서 sort는 내부적으로 추가적인 메모리를 사용해서 공간복잡도가 O(1)은 아닐 것 같습니다! There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 엇 그렇네요 정렬할 때 공간복잡도가 On 사용됩니다! |
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nums.sort((a,b) => a - b); | ||
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for (let i = 0 ; i <= nums.length; i++) { | ||
if (i !== nums[i]) return i | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 숫자가 0부터 시작하니까 이렇게 비교해서 풀어도 좋겠네요! 배워갑니다 :) |
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} | ||
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}; | ||
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console.log(missingNumber([3, 0, 1])) | ||
console.log(missingNumber([0, 1])) | ||
console.log(missingNumber([9,6,4,2,3,5,7,0,1])) |
Original file line number | Diff line number | Diff line change |
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// 시간복잡도: O(n) | ||
// 공간복잡도: O(n) | ||
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/** | ||
* @param {string} s | ||
* @return {boolean} | ||
*/ | ||
var isPalindrome = function(s) { | ||
const strs = s.replace(/[^a-z0-9]/gi, '').toLowerCase(); | ||
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let leftIdx = 0; | ||
let rightIdx = strs.length - 1 | ||
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while (leftIdx <= rightIdx) { | ||
if (strs[leftIdx] !== strs[rightIdx]) return false | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 오 양쪽에서 인덱스를 증가시키거나 감소시키면서 하나씩 비교하니 풀이가 굉장히 깔끔하네요! |
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leftIdx++ | ||
rightIdx-- | ||
} | ||
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return true | ||
}; | ||
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const s = "A man, a plan, a canal: Panama" | ||
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console.log(isPalindrome(s)) | ||
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// m은 board의 행 수, n은 board의 열 수, 4(상하좌우), l(word) | ||
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// 시간복잡도: O(m * n * 4L) | ||
// 공간복잡도: O(m * n + L) | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. FYI #432 (comment) |
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const d_row = [1, -1, 0, 0] | ||
const d_col = [0, 0, -1, 1] | ||
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const isMoveBoard = (new_row, new_col, board) => { | ||
return new_row >= 0 && new_row < board.length && new_col >= 0 && new_col < board[0].length | ||
} | ||
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/** | ||
* @param {character[][]} board | ||
* @param {string} word | ||
* @return {boolean} | ||
*/ | ||
var exist = function(board, word) { | ||
let result = false | ||
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const visited = Array.from({length: board.length} , () => Array.from({length: board[0].length}).fill(false)) | ||
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const dfs = (strs, row, col, count) => { | ||
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if (strs[count] !== word[count]) return | ||
if (strs.length > word.length) return | ||
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if (strs === word) { | ||
result = true | ||
return | ||
} | ||
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for (let i = 0 ; i < d_row.length; i++) { | ||
const new_row = row + d_row[i] | ||
const new_col = col + d_col[i] | ||
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if (isMoveBoard(new_row, new_col, board) && visited[new_row][new_col] !== true){ | ||
visited[new_row][new_col] = true | ||
dfs(strs+board[new_row][new_col], new_row, new_col, count+1) | ||
visited[new_row][new_col] = false | ||
} | ||
} | ||
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} | ||
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for (let row = 0 ; row < board.length ; row++) { | ||
for (let col = 0 ; col < board[0].length ; col++) { | ||
visited[row][col] = true | ||
dfs(board[row][col], row, col , 0) | ||
visited[row][col] = false | ||
} | ||
} | ||
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return result | ||
}; | ||
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console.log(exist([["A","B","C","E"], | ||
["S","F","C","S"], | ||
["A","D","E","E"]], "ABCCED")) | ||
console.log(exist([["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], "SEE")) | ||
console.log(exist([["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], "ABCB")) | ||
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엇 그런데 문제 조건에서 O(n)에 동작하는 코드를 작성해야 한다고 했는데, 해당 코드는 시간복잡도가 nlogn이 되는 것 같습니다!
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앗 그냥 생각나는데로 풀어버렸네요 ㅜ
정렬을 안쓰고 풀어봐야겠어요 ~