Leetcode48. 旋转图像_leetcode48-旋转图像-优快云博客

本文链接：https://blog.youkuaiyun.com/sunhaiting666/article/details/104885314

本文详细解析了LeetCode第48题“旋转图像”的解决方案，介绍了三种不同的算法实现，包括矩阵转置加翻转、旋转以及单次循环方法，并提供了Scala和Java代码示例。

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Leetcode48. 旋转图像

题目：
给定一个 n × n 的二维矩阵表示一个图像。
将图像顺时针旋转 90 度。
说明：
你必须在原地旋转图像，这意味着你需要直接修改输入的二维矩阵。请不要使用另一个矩阵来旋转图像。

示例 1:

给定 matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

原地旋转输入矩阵，使其变为:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

示例 2:

给定 matrix =
[
  [ 5, 1, 9,11],
  [ 2, 4, 8,10],
  [13, 3, 6, 7],
  [15,14,12,16]
], 

原地旋转输入矩阵，使其变为:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

题解：
1.转置+翻转
2。旋转
scala代码：

/**
    * 转置+翻转
    *
    * @param matrix
    */
  def rotate1(matrix: Array[Array[Int]]): Unit = {

    val n = matrix.length
    //矩阵转置
    for (i <- 0 until n) {
      for (j <- i until n) {
        val temp = matrix(j)(i)
        matrix(j)(i) = matrix(i)(j)
        matrix(i)(j) = temp
      }
    }
    //翻转
    for (i <- 0 until n) {
      for (j <- 0 until n / 2) {
        val temp = matrix(i)(j)
        matrix(i)(j) = matrix(i)(n - j - 1)
        matrix(i)(n - j - 1) = temp
      }
    }
  }

  /**
    * 旋转
    *
    * @param matrix
    */
  def rotate2(matrix: Array[Array[Int]]): Unit = {

    val n = matrix.length
    for (i <- 0 until (n / 2 + n % 2)) {
      for (j <- 0 until n / 2) {
        val nums = new Array[Int](4)
        var row = i
        var col = j
        for (i <- 0 until 4) {
          nums(i) = matrix(row)(col)
          val temp = row
          row = col
          col = n - 1 - temp
        }
        for (i <- 0 until 4) {
          matrix(row)(col) = nums((i + 3) % 4)
          val temp = row
          row = col
          col = n - 1 - temp
        }
      }
    }
  }

  /**
    * 单次循环
    *
    * @param matrix
    */
  def rotate3(matrix: Array[Array[Int]]): Unit = {
    val n = matrix.length
    for (i <- 0 until (n + 1) / 2) {
      for (j <- 0 until n / 2) {
        val temp = matrix(n - 1 - j)(i)
        matrix(n - 1 - j)(i) = matrix(n - 1 - i)(n - j - 1)
        matrix(n - 1 - i)(n - j - 1) = matrix(j)(n - 1 - i)
        matrix(j)(n - 1 - i) = matrix(i)(j)
        matrix(i)(j) = temp
      }
    }
  }

java代码：

public static int[][] rotate(int[][] matrix) {

        int n = matrix.length;
        for (int i = 0; i < (n + 1) / 2; i++) {
            for (int j = 0; j < n / 2; j++) {
                int tmp = matrix[j][n - i - 1];
                matrix[j][n - i - 1] = matrix[i][j];
                matrix[i][j] = matrix[n - j - 1][i];
                matrix[n - j - 1][i] = matrix[n - i - 1][n - j - 1];
                matrix[n - i - 1][n - j - 1] = tmp;
            }
        }
        return matrix;
    }