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  • /**
    
     Given a matrix A, return the transpose of A.
    
     The transpose of a matrix is the matrix flipped over it's main diagonal, switching the row and column indices of the matrix.
    
    
    
     Example 1:
    
     Input: [[1,2,3],[4,5,6],[7,8,9]]
     Output: [[1,4,7],[2,5,8],[3,6,9]]
     Example 2:
    
     Input: [[1,2,3],[4,5,6]]
     Output: [[1,4],[2,5],[3,6]]
    
    
     Note:
    
     1 <= A.length <= 1000
     1 <= A[0].length <= 1000
    
     */
    public class Transpose_Matrix {
    
        class Solution {
            public int[][] transpose(int[][] A) {
    
                if (A == null || A.length == 0) {
                    return A;
                }
    
                int m = A.length;
                int n = A[0].length;
    
                int[][] transPos = new int[n][m];
    
                for (int i = 0; i < m; i++) {
                    for (int j = 0; j < n; j++) {
                        transPos[j][i] = A[i][j];
                    }
                }
    
                return transPos;
            }
        }
    }
    
    ############
    
    class Solution {
        public int[][] transpose(int[][] matrix) {
            int m = matrix.length, n = matrix[0].length;
            int[][] ans = new int[n][m];
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < m; ++j) {
                    ans[i][j] = matrix[j][i];
                }
            }
            return ans;
        }
    }
    
  • // OJ: https://leetcode.com/problems/transpose-matrix/
    // Time: O(MN)
    // Space: O(1) extra space
    class Solution {
    public:
        vector<vector<int>> transpose(vector<vector<int>>& A) {
            int M = A.size(), N = A[0].size();
            vector<vector<int>> ans(N, vector<int>(M));
            for (int i = 0; i < M; ++i) {
                for (int j = 0; j < N; ++j) {
                    ans[j][i] = A[i][j];
                }
            }
            return ans;
        }
    };
    
  • class Solution:
        def transpose(self, matrix: List[List[int]]) -> List[List[int]]:
            return list(zip(*matrix))
    
    ############
    
    class Solution:
        def transpose(self, A):
            """
            :type A: List[List[int]]
            :rtype: List[List[int]]
            """
            rows, cols = len(A), len(A[0])
            res = [[0] * rows for _ in range(cols)]
            for row in range(rows):
                for col in range(cols):
                    res[col][row] = A[row][col]
            return res
    
  • func transpose(matrix [][]int) [][]int {
    	m, n := len(matrix), len(matrix[0])
    	ans := make([][]int, n)
    	for i := range ans {
    		ans[i] = make([]int, m)
    		for j := range ans[i] {
    			ans[i][j] = matrix[j][i]
    		}
    	}
    	return ans
    }
    
  • /**
     * @param {number[][]} matrix
     * @return {number[][]}
     */
    var transpose = function (matrix) {
        const m = matrix.length;
        const n = matrix[0].length;
        const ans = new Array(n).fill(0).map(() => new Array(m).fill(0));
        for (let i = 0; i < n; ++i) {
            for (let j = 0; j < m; ++j) {
                ans[i][j] = matrix[j][i];
            }
        }
        return ans;
    };
    
    
  • class Solution {
        public int[][] transpose(int[][] A) {
            int rows = A.length, columns = A[0].length;
            int[][] transpose = new int[columns][rows];
            for (int i = 0; i < rows; i++) {
                for (int j = 0; j < columns; j++)
                    transpose[j][i] = A[i][j];
            }
            return transpose;
        }
    }
    
    ############
    
    class Solution {
        public int[][] transpose(int[][] matrix) {
            int m = matrix.length, n = matrix[0].length;
            int[][] ans = new int[n][m];
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < m; ++j) {
                    ans[i][j] = matrix[j][i];
                }
            }
            return ans;
        }
    }
    
  • // OJ: https://leetcode.com/problems/transpose-matrix/
    // Time: O(MN)
    // Space: O(1) extra space
    class Solution {
    public:
        vector<vector<int>> transpose(vector<vector<int>>& A) {
            int M = A.size(), N = A[0].size();
            vector<vector<int>> ans(N, vector<int>(M));
            for (int i = 0; i < M; ++i) {
                for (int j = 0; j < N; ++j) {
                    ans[j][i] = A[i][j];
                }
            }
            return ans;
        }
    };
    
  • class Solution:
        def transpose(self, matrix: List[List[int]]) -> List[List[int]]:
            return list(zip(*matrix))
    
    ############
    
    class Solution:
        def transpose(self, A):
            """
            :type A: List[List[int]]
            :rtype: List[List[int]]
            """
            rows, cols = len(A), len(A[0])
            res = [[0] * rows for _ in range(cols)]
            for row in range(rows):
                for col in range(cols):
                    res[col][row] = A[row][col]
            return res
    
  • func transpose(matrix [][]int) [][]int {
    	m, n := len(matrix), len(matrix[0])
    	ans := make([][]int, n)
    	for i := range ans {
    		ans[i] = make([]int, m)
    		for j := range ans[i] {
    			ans[i][j] = matrix[j][i]
    		}
    	}
    	return ans
    }
    
  • /**
     * @param {number[][]} matrix
     * @return {number[][]}
     */
    var transpose = function (matrix) {
        const m = matrix.length;
        const n = matrix[0].length;
        const ans = new Array(n).fill(0).map(() => new Array(m).fill(0));
        for (let i = 0; i < n; ++i) {
            for (let j = 0; j < m; ++j) {
                ans[i][j] = matrix[j][i];
            }
        }
        return ans;
    };
    
    

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