Leetcode [Easy] 48 - Rotate Image
LeetCode 48
Rotate Image
문제
You are given an n x n 2D matrix representing an image, rotate the image by 90 degrees (clockwise).
You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.
Example 1:
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [[7,4,1],[8,5,2],[9,6,3]]
Example 2:
Input: matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]
Output: [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]
Example 3:
Input: matrix = [[1]]
Output: [[1]]
Example 4:
Input: matrix = [[1,2],[3,4]]
Output: [[3,1],[4,2]]
Constraints:
matrix.length == n
matrix[i].length == n
1 <= n <= 20
-1000 <= matrix[i][j] <= 1000
나의 코드
/**
Do not return anything, modify matrix in-place instead.
*/
function rotate(matrix: number[][]): void {
interface Memo {
[index: number]: number[];
}
const memo: Memo = {};
for (let i: number = 0; i < matrix.length; i++) {
memo[i] = [...matrix[i]];
}
for (let i: number = 0; i < matrix.length; i++) {
for (let j: number = 0; j < matrix[i].length; j++) {
matrix[i][j] = memo[matrix.length - j - 1][i];
}
}
}
참고 코드
var rotate = function (matrix) {
// Transpose the matrix
for (let i = 0; i < matrix.length; i++) {
for (let j = i; j < matrix[0].length; j++) {
let temp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = temp;
}
}
// Reverse each row
for (let i = 0; i < matrix.length; i++) {
for (let j = 0; j < matrix[0].length / 2; j++) {
let temp = matrix[i][j];
matrix[i][j] = matrix[i][matrix[0].length - j - 1];
matrix[i][matrix[0].length - j - 1] = temp;
}
}
};
배운점
- 배열 이동을 다룰 때 꼭 memo 를 사용하는 방법 말고도 방법이 있다.
- 참고 코드의 방법 잘 기억해두기
(i, j 를 역으로 뒤집고 각 열별로 reverse 해주니 90도 회전이 되었다.)
Subscribe via RSS