Question

867. 转置矩阵

English Version

题目描述

给你一个二维整数数组 matrix， 返回 matrix 的 转置矩阵 。

矩阵的 转置 是指将矩阵的主对角线翻转，交换矩阵的行索引与列索引。

 

示例 1：

输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
输出：[[1,4,7],[2,5,8],[3,6,9]]

示例 2：

输入：matrix = [[1,2,3],[4,5,6]]
输出：[[1,4],[2,5],[3,6]]

 

提示：

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 1000
• 1 <= m * n <= 105
• -109 <= matrix[i][j] <= 109

解法

方法一

class Solution:
  def transpose(self, matrix: List[List[int]]) -> List[List[int]]:
    return list(zip(*matrix))

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;
  }
}

class Solution {
public:
  vector<vector<int>> transpose(vector<vector<int>>& matrix) {
    int m = matrix.size(), n = matrix[0].size();
    vector<vector<int>> ans(n, vector<int>(m));
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < m; ++j)
        ans[i][j] = matrix[j][i];
    return ans;
  }
};

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;
};