[LeetCode] Summary of Matrix Simulation Related Topics

A summary of matrix simulation related topics

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59.螺旋矩阵 II

59. 螺旋矩阵 II

给你一个正整数 n ,生成一个包含 1 到 n2 所有元素,且元素按顺时针顺序螺旋排列的 n x n 正方形矩阵 matrix .

示例 1：

输入：n = 3
输出：[[1,2,3],[8,9,4],[7,6,5]]

示例 2：

输入：n = 1
输出：[[1]]

题解思路

The idea here is just transport,The original link is below

• 生成一个 n×n 空矩阵 dep,随后模拟整个向内环绕的填入过程：
  • 定义当前左右上下边界 left,right,top,bottom,初始值 count,迭代终止值 total = n * n;
  • 当 count < total 时,始终按照 从左到右 从上到下 从右到左 从下到上 填入顺序循环,每次填入后：
    • 执行 ++count：得到下一个需要填入的数字;
    • 更新边界：例如从左到右填完后,上边界 top ++,相当于上边界向内缩 1.
• 最终返回 mat 即可.

原文链接：Spiral Matrix II （模拟法,设定边界,代码简短清晰）

var generateMatrix = function(n) {
    
    //四个边界
    let left = 0;
    let right = n - 1;
    let bottom = n - 1;
    let top = 0;
    const total = n * n;//矩阵总数
    const dep = [];//初始化矩阵
    for(let i = 0; i < n; i++){
    
        dep[i] = [];
    }
    let count = 0;
    while(count < total){
    
        for(let i = left; i <= right; i++) dep[top][i] = ++count;//从左到右
        top++;
        for(let i = top; i <= bottom; i++) dep[i][right] = ++count;//从上到下
        right--;
        for(let i = right; i >= left; i--) dep[bottom][i] = ++count;//从右到左
        bottom--;
        for(let i = bottom; i >= top; i--) dep[i][left] = ++count;//从下到上
        left++;
    }
    return dep;
};

54. 螺旋矩阵

54. 螺旋矩阵

给你一个 m 行 n 列的矩阵 matrix ,请按照 顺时针螺旋顺序 ,返回矩阵中的所有元素.

示例 1：

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

示例 2：

输入：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
输出：[1,2,3,4,8,12,11,10,9,5,6,7]

题解思路

• When iterating over all items,res The array is constructed.我们可以用 res 数组的长度 等于 The number of entries in the matrix,作为循环的结束条件
• Does not mean to continue to traverse,等于就 break

举例说明

• 每遍历一条边,The starting point of the next edge traversal is “挤占”,To update the corresponding boundaries
• 注意到,Possibly in the loop,Suddenly the conditions of the loop are no longer satisfied,即top > bottom或left > right,其中一对边界彼此交错了
• This means that all items have been traversed,要break了,如果没有及时 break ,就会重复遍历

For the matrix on the left in the image above：
The four boundary values ​​of the upper, lower, left and right are initialized respectively：

• top = 0, right = 5, bottom = 4, left = 0

When the outermost circle is traversed,The four boundary values ​​are now ：

• top = 1, right = 4, bottom = 3, left = 1

When the second traversal is completed,The four boundary values ​​are now ：

• top = 2, right = 3, bottom = 2, left = 2

At this point, it starts to traverse the middle two numbers： 15,16,After traversing from left to right,此时执行top++,那么此时 top=3.

If not end the loop,Then the next value will overwrite the value already assigned in the picture22

So the loop should be exited at this point

解决办法

• 每遍历完一条边,After updating the corresponding boundaries,都加上一句if (top > bottom || left > right) break;,Avoid repeated traversal due to not exiting the loop in time.
• 且,“遍历完成”这个时间点,Either happens at the end of the traversal“上边”,Either happens at the end of the traversal“右边”
• So just after these two steps,加 if (top > bottom || left > right) break 即可

var spiralOrder = function (matrix) {
    
  if (matrix.length == 0) return []
  const res = []
  let top = 0, bottom = matrix.length - 1, left = 0, right = matrix[0].length - 1
  const size = matrix.length * matrix[0].length
  while (res.length !== size) {
     // The traversal is not over yet
    for (let i = left; i <= right; i++) res.push(matrix[top][i])
    top++
    for (let i = top; i <= bottom; i++) res.push(matrix[i][right])
    right--
    if (res.length === size) break // 遍历结束
    for (let i = right; i >= left; i--)  res.push(matrix[bottom][i])
    bottom--
    for (let i = bottom; i >= top; i--) res.push(matrix[i][left])
    left++
  }
  return res
};

剑指 Offer 29. 顺时针打印矩阵

剑指 Offer 29. 顺时针打印矩阵

输入一个矩阵,按照从外向里以顺时针的顺序依次打印出每一个数字.

示例 1：

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

示例 2：

输入：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
输出：[1,2,3,4,8,12,11,10,9,5,6,7]

题解思路

• 这个题与54题类似

var spiralOrder = function(matrix) {
    
    if(matrix.length === 0){
    
        return [];
    }
    let res = [];
    const total = matrix.length * matrix[0].length;
    let top = 0, right = matrix[0].length - 1, bottom = matrix.length - 1, left = 0;
    while(res.length !== total){
    
        for(let i = left; i <= right; i++) res.push(matrix[top][i]);
        top++;
        for(let i = top; i <= bottom; i++) res.push(matrix[i][right]);
        right--;
        if(res.length === total) break;
        for(let i = right; i >= left; i--) res.push(matrix[bottom][i]);
        bottom--;
        for(let i = bottom; i >= top; i--) res.push(matrix[i][left]);
        left++;
    }
    return res;
};

48. 旋转图像

48. 旋转图像

给定一个 n × n 的二维矩阵 matrix 表示一个图像.请你将图像顺时针旋转 90 度.

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.

题解思路

作为例子,先将其通过水平轴翻转得到：

再根据主对角线翻转得到：

就得到了答案.这是为什么呢？对于水平轴翻转而言,We just need to enumerate the elements in the upper half of the matrix,Swap with the elements in the lower half,即

var rotate = function(matrix) {
    
    const n = matrix.length;
    //水平中轴线翻转
    for(let i = 0; i < Math.floor(n/2); i++){
    
        for(let j = 0; j < n; j++){
    
            [matrix[i][j], matrix[n - i - 1][j]] = [matrix[n - i - 1][j], matrix[i][j]];
        }
    }
    //主对角线翻转
    for(let i = 0; i < n; i++){
    
        for(let j = 0; j < i; j++){
    
            [matrix[i][j], matrix[j][i]] = [matrix[j][i], matrix[i][j]];
        }
    }
};

There are other topics similar to this one：面试题 01.07. 旋转矩阵

做法相同,在这里不再赘述\

566. 重塑矩阵

566. 重塑矩阵

在 MATLAB 中,有一个非常有用的函数 reshape ,它可以将一个 m x n 矩阵重塑为另一个大小不同（r x c）的新矩阵,但保留其原始数据.

给你一个由二维数组 mat 表示的 m x n 矩阵,以及两个正整数 r 和 c ,分别表示想要的重构的矩阵的行数和列数.

重构后的矩阵需要将原始矩阵的所有元素以相同的 行遍历顺序 填充.

如果具有给定参数的 reshape 操作是可行且合理的,则输出新的重塑矩阵;否则,输出原始矩阵.

题解思路

• Convert the original two-dimensional array to one-dimensional
• Then judge whether the length matches
• Fill at the end

var matrixReshape = function(mat, r, c) {
    
    const newMat = [];
    //将二维数组转换为一维数组
    for(let i = 0; i < mat.length; i++){
    
        newMat.push(...mat[i]);
    }
    //Determine whether the length meets the requirements
    if(r * c !== newMat.length) return mat;
    
    //重塑矩阵
    for(let i = 0; i < r; i++){
    
        const item = [];
        //每行c个
        for(let j = 0; j < c; j++){
    
            //将c个元素从头部拿出,并放入暂存的item数组
            item.push(newMat.shift());
        }
        //当前行收集完毕,Push to the end of the new array
        newMat.push(item);
    }
    return newMat;
};

注意：JS There is no concept of a two-dimensional array in ,You need to create a two-dimensional array object yourself
首先JSThe way to create a 2D array is not straightforwardnew Array[][],
而是要先new一个一维数组,Then put each element of this one-dimensional array new 为一个数组

    const dep = [];//初始化矩阵
    for(let i = 0; i < n; i++){
    
        dep[i] = [];
    }

Or write each row as an array,Refer to this topic for specific code