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luzhipeng feat: #29 #48 #49 #56 #98 aea6437 May 29, 2019
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题目地址

https://leetcode.com/problems/rotate-image/description/

题目描述

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Note:

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:

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

rotate the input matrix in-place such that it becomes:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]
Example 2:

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

rotate the input matrix in-place such that it becomes:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

思路

这道题目让我们 in-place,也就说空间复杂度要求 O(1),如果没有这个限制的话,很简单。

通过观察发现,我们只需要将第 i 行变成第 n - i - 1 列, 因此我们只需要保存一个原有矩阵,然后按照这个规律一个个更新即可。

48.rotate-image-1

代码:

var rotate = function(matrix) {
  // 时间复杂度O(n^2) 空间复杂度O(n)
  const oMatrix = JSON.parse(JSON.stringify(matrix)); // clone
  const n = oMatrix.length;
  for (let i = 0; i < n; i++) {
    for (let j = 0; j < n; j++) {
      matrix[j][n - i - 1] = oMatrix[i][j];
    }
  }
};

如果要求空间复杂度是O(1)的话,我们可以用一个temp记录即可,这个时候就不能逐个遍历了。 比如遍历到1的时候,我们把1存到temp,然后更新1的值为7。 1被换到了3的位置,我们再将3存到temp,依次类推。 但是这种解法写起来比较麻烦,这里我就不写了。

事实上有一个更加巧妙的做法,我们可以巧妙地利用对称轴旋转达到我们的目的,如图,我们先进行一次以对角线为轴的翻转,然后 再进行一次以水平轴心线为轴的翻转即可。

48.rotate-image-2

这种做法的时间复杂度是O(n^2) ,空间复杂度是O(1)

关键点解析

  • 矩阵旋转操作

代码

/*
 * @lc app=leetcode id=48 lang=javascript
 *
 * [48] Rotate Image
 */
/**
 * @param {number[][]} matrix
 * @return {void} Do not return anything, modify matrix in-place instead.
 */
var rotate = function(matrix) {
  // 时间复杂度O(n^2) 空间复杂度O(1)

  // 做法: 先沿着对角线翻转,然后沿着水平线翻转
  const n = matrix.length;
  function swap(arr, [i, j], [m, n]) {
    const temp = arr[i][j];
    arr[i][j] = arr[m][n];
    arr[m][n] = temp;
  }
  for (let i = 0; i < n - 1; i++) {
    for (let j = 0; j < n - i; j++) {
      swap(matrix, [i, j], [n - j - 1, n - i - 1]);
    }
  }

  for (let i = 0; i < n / 2; i++) {
    for (let j = 0; j < n; j++) {
      swap(matrix, [i, j], [n - i - 1, j]);
    }
  }
};
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