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[gitsunmin] Week 09 Solutions #527
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/** | ||
* https://leetcode.com/problems/find-minimum-in-rotated-sorted-array | ||
* time complexity : O(log n) | ||
* space complexity : O(1) | ||
*/ | ||
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function findMin(nums: number[]): number { | ||
let left = 0; | ||
let right = nums.length - 1; | ||
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while (left < right) { | ||
const mid = Math.floor((left + right) / 2); | ||
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if (nums[mid] > nums[right]) left = mid + 1; | ||
else right = mid; | ||
} | ||
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return nums[left]; | ||
}; |
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/** | ||
* https://leetcode.com/problems/linked-list-cycle/ | ||
* time complexity : O(n) | ||
* space complexity : O(1) | ||
*/ | ||
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export class ListNode { | ||
val: number | ||
next: ListNode | null | ||
constructor(val?: number, next?: ListNode | null) { | ||
this.val = (val === undefined ? 0 : val) | ||
this.next = (next === undefined ? null : next) | ||
} | ||
} | ||
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function hasCycle(head: ListNode | null): boolean { | ||
if (!head || !head.next) { | ||
return false; | ||
} | ||
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let slow: ListNode | null = head; | ||
let fast: ListNode | null = head.next; | ||
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while (slow !== fast) { | ||
if (!fast || !fast.next) return false; | ||
slow = slow!.next; | ||
fast = fast.next.next; | ||
} | ||
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return true; | ||
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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. 코드 리뷰를 하면서 드는 생각인데 union-find로도 풀어볼 걸 그랬네요 |
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}; |
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/** | ||
* https://leetcode.com/problems/two-sum | ||
* time complexity : O(m x m) | ||
* space complexity : O(m x n) | ||
*/ | ||
function pacificAtlantic(heights: number[][]): number[][] { | ||
const m = heights.length; | ||
const n = heights[0].length; | ||
const pacific: boolean[][] = Array.from({ length: m }, () => Array(n).fill(false)); | ||
const atlantic: boolean[][] = Array.from({ length: m }, () => Array(n).fill(false)); | ||
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const directions = [[1, 0], [-1, 0], [0, 1], [0, -1]]; | ||
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function dfs(r: number, c: number, visited: boolean[][], prevHeight: number) { | ||
if (r < 0 || c < 0 || r >= m || c >= n || visited[r][c] || heights[r][c] < prevHeight) { | ||
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. 저도 귀찮아서 한 줄로 쓰기는 하는데, 리뷰어의 입장에서 코드 스멜이나 단축 평가를 고려하면 적절한 조건들로 나누는 게 좋을 것 같습니다. 이 경우는 index error가 일어나지 않을 r, c를 한정하기 위해 순서가 어쩔 수 없지만... 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. @lymchgmk |
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return; | ||
} | ||
visited[r][c] = true; | ||
for (const [dr, dc] of directions) { | ||
dfs(r + dr, c + dc, visited, heights[r][c]); | ||
} | ||
} | ||
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for (let i = 0; i < m; i++) { | ||
dfs(i, 0, pacific, heights[i][0]); | ||
dfs(i, n - 1, atlantic, heights[i][n - 1]); | ||
} | ||
for (let i = 0; i < n; i++) { | ||
dfs(0, i, pacific, heights[0][i]); | ||
dfs(m - 1, i, atlantic, heights[m - 1][i]); | ||
} | ||
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const result: number[][] = []; | ||
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for (let r = 0; r < m; r++) { | ||
for (let c = 0; c < n; c++) { | ||
if (pacific[r][c] && atlantic[r][c]) { | ||
result.push([r, c]); | ||
} | ||
} | ||
} | ||
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return result; | ||
} |
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굳이 floor은 안사용하셔도 괜찮을 것 같습니다. 이분탐색에서 while문에 등호를 넣거나 if나 else에서 등호를 넣거나 테스트 해보시면 좋을 것 같습니다.