[Vssue]0059.螺旋矩阵II.md

[youngyangyang04]

https://programmercarl.com/0059.%E8%9E%BA%E6%97%8B%E7%9F%A9%E9%98%B5II.html

[hsienshaun]

我用gpt生成的, 稍微简洁一些, 模拟过程也好理解一点...

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        vector<vector<int>> Matrix(n, vector<int>(n, 0));
        int top = 0, bottom = n - 1, left = 0, right = n-1;
        int num = 1;

        while(num <= n * n){
            for (int i = left; i<= right; ++i){
                Matrix[top][i] = num ++;

            }
            ++top ;

            for (int i = top; i <= bottom; ++i){
                Matrix[i][right] = num ++;
            }
            --right ;

            for (int i = right; i >= left; --i){
                Matrix[bottom][i] = num++;
            }
            --bottom;
            for (int i = bottom; i >= top; --i){
                Matrix[i][left] = num++;
            }
            ++left;
        }
        return Matrix;
    }
};

[superGinga]

理解了老师的解法之后，自己闭卷手动写了一遍，其实思路不难就是把手工画圈的过程模拟为程序运行的过程，最关键的地方当然就在于始终都要保证区间一致性，不然容易把自己绕晕。 另外，第三个和第四个for循环，虽然是j--和i--的过程，但不必拘泥于左右区间的开闭，保证指针运动的“起始区间”为闭，“结束区间”为开即可。 class Solution { public: vector<vector> generateMatrix(int n) { vector<vector> nums(n,vector(n,0)); int loop=n; int i=0,j=0,count=1,startx=0,starty=0; while(loop--) { i=starty,j=startx; for(;j<loop;j++) { nums[i][j]=count++; } for(;i<loop;i++) { nums[i][j]=count++; } for(;j>startx;j--) { nums[i][j]=count++; } for(;i>starty;i--) { nums[i][j]=count++; } startx++;starty++; } int mid=n/2; if(n%2==1)nums[mid][mid]=count; return nums; } };

[wing2791]

思路应该是一样的，不过我把他抽象成了函数的形式，没考虑开闭区间 C++

class Solution {
public:
    int index = 1;
    void horizontal(bool isRight,vector<vector<int>> & res,int row,int n){
        if(isRight){
            for(int i = 0;i<n;i++){
                if(res[row][i] == 0) res[row][i] = index++;
            }
        }else{
            for(int i = n-1;i>-1;i--){
                if(res[row][i] == 0) res[row][i] = index++;
            }
        }
    }

    void vertical(bool isDown,vector<vector<int>> & res,int col,int n){
        if(isDown){
            for(int i = 0;i<n;i++){
                if(res[i][col] == 0) res[i][col] = index++;
            }
        }else{
            for(int i = n-1;i>-1;i--){
                if(res[i][col] == 0) res[i][col] = index++;
            }
        }
    }

    vector<vector<int>> generateMatrix(int n) {
        vector<vector<int>> res(n,vector<int>(n));
        int maxN = (n+1)/2;
        for(int i = 0;i<maxN;i++){
            horizontal(true,res,i,n);
            vertical(true,res,n-1-i,n);
            horizontal(false,res,n-1-i,n);
            vertical(false,res,i,n);
        }
        return res;
    }
};

[Chenromeo]

java的 稍微简化了一下，把各种中间变量抽象成turn

class Solution {
    public int[][] generateMatrix(int n) {
        int[][] matrix = new int[n][n];
        int count = 1;  // count 记录需要填的值
        int turn = 1; // turn记录当前转圈轮数 
        int x = 0, y = 0; // 初始化数组坐标
        while(turn <= n/2){ // 当 当前轮数turn <= n/2（每轮按1/2缩减），按左闭右开赋值，即[turn, n-turn-1)
            x = turn - 1; 
            y = turn - 1; // 初始化每一轮转圈的起始坐标（x++, y++）
            for (; y < n - turn; y++) { // 转圈赋值的上边，
                matrix[x][y] = count++;
            }
            for (; x < n - turn; x++) { // 转圈赋值的右边
                matrix[x][y] = count++;
            }
            for (; y > turn - 1; y--) { // 转圈赋值的下边
                matrix[x][y] = count++;
            }
            for (; x > turn - 1; x--) { // 转圈赋值的左边
                matrix[x][y] = count++;
            }
            turn++;
        }
        if (n % 2 == 1) { // n为奇数，需单独处理中心位置
                matrix[turn - 1][turn - 1] = count;
            }
        return matrix;
    }
}

[Du1in9]

class Solution {
    public int[][] generateMatrix(int n) {
        int [][] nums = new int[n][n];
        int startX = 0, startY = 0;
        int loop = 1, count = 1, offset = 1;
        int i, j;
        while(loop <= n / 2){
            i = startX; j = startY;
            for(; j < n - offset; j++){ // 1. n - offset, not n
                nums[i][j] = count++;
            }
            for(; i < n - offset; i++){
                nums[i][j] = count++;
            }
            for(; j >= offset; j--){    // 2. offset, not 0
                nums[i][j] = count++;
            }
            for(; i >= offset; i--){
                nums[i][j] = count++;
            }
            startX++; startY++; offset++; loop++;
        }
        if(n % 2 == 1){     // n 为奇数时, 单独处理矩阵中心的值
            nums[startX][startY] = count;
        }
        return nums;
    }
}

[haveadrame]

打卡

[casperking666]

为啥会有[x][y]这样的表示方式？太奇怪了，从几何想怎么都是[y][x]

[Quiet0214]

好复杂啊😵

[guangjingchao]

文中Java写法处理n为奇数的if语句应该放在while循环外面才能处理n=1的情况，代码如下：class Solution { public int[][] generateMatrix(int n) { int count = 1; int i, j; int loop = 1; int startX = 0, startY = 0; int offSet = 1; int[][] arrs = new int[n][n]; while (loop <= n / 2) {//左闭右开 //上 for (j = startY; j < n - offSet; j++) { //<= 与 <，前者数组越界 arrs[startX][j] = count++; }

        //右
        for (i = startX; i < n - offSet; i++) {
            arrs[i][j] = count++;
        }

        //下
        for (; j > startY; j--) {  //>startY
            arrs[i][j] = count++;
        }

        //左
        for (; i > startX; i--) { //>startX
            arrs[i][j] = count++;
        }

        offSet++;
        startX++;
        startY++;
        loop++;

    }
     if (n % 2 == 1) { //放在外面，处理n=1的情况
            arrs[startX][startY] = count;
        }
    return arrs;
}

}

[Ding-Jiaxiong]

打卡

    public static int[][] generateMatrix(int n) {

        int[][] res = new int[n][n]; // 结果二维数组

        int left = 0; // 左边界
        int right = n - 1;  // 右边界
        int top = 0; // 上边界
        int bottom = n - 1; // 下边界

        int startnum = 1; // 从 1 开始填
        int endnum = n * n;  // 填充结束值

        while (startnum <= endnum) {

            for (int i = left; i <= right; i++) { // 1. 从左到右填充上行
                res[top][i] = startnum; // 行不变，一直是上边界
                startnum++;  // 值自增
            }

            top++; // 一行填充完毕

            for (int i = top; i <= bottom; i++) { // 2. 从上到下填充右列
                res[i][right] = startnum;  // 列不变，一直是右边界
                startnum++;

            }

            right--; // 一列填充完毕

            for (int i = right; i >= left; i--) { // 3. 从右到左填充下行
                res[bottom][i] = startnum; // 行不变，一直是下边界
                startnum++;

            }

            bottom--; // 一行填充完毕

            for (int i = bottom; i >= top; i--) { // 4. 从下到上填充左列
                res[i][left] = startnum; // 列不变，一直是左边界
                startnum++;

            }

            left++; // 一列完毕

        }

        return res;
    }

[Rafael8830]

0802打卡

[JIE-yx]

不如我这个 class Solution {

private int fillNum = 0; 

private int[][] nextStep = new int[4][2];

public int[][] generateMatrix(int n) {
    int[][] matrix = new int[n][n];
    for (int i = 0; i < n; i ++) {
        matrix[0][i] = i + 1;
        fillNum++;
    }
    buildNextStep();
    int startRow = 0;
    int startCol = n - 1;
    int direction = 1;
    while (fillNum != n * n) {
        int[] nextStarts=fill(matrix, direction, startRow, startCol);
        direction++;
        if (direction > 3) {
            direction = 0;
        }
        startRow = nextStarts[0];
        startCol = nextStarts[1];
    }
    return matrix;
}

private void buildNextStep() {
    // 右，下，左，上，右，下，左，上
    nextStep[0][0] = 0;
    nextStep[0][1] = 1;
    nextStep[1][0] = 1;
    nextStep[1][1] = 0;
    nextStep[2][0] = 0;
    nextStep[2][1] = -1;
    nextStep[3][0] = -1;
    nextStep[3][1] = 0;
}

private int[] fill(int[][] matrix, int direction, int startRow, int startCol) {
    int num = matrix[startRow][startCol];
    num++;
    int n = matrix.length;
    int i = startRow + nextStep[direction][0];
    int j = startCol + nextStep[direction][1];
    while (i >= 0 && i < n && j >= 0 && j < n && matrix[i][j] == 0) {
        matrix[i][j] = num;
        num++;
        i = i + nextStep[direction][0];
        j = j + nextStep[direction][1];
        fillNum++;
    }
    return new int[]{i - nextStep[direction][0], j - nextStep[direction][1]};
}

}

[Bryan-CR7]

打卡：大循环条件中，若循环条件为count <= n, 无法规避n=1无法执行的情况，而loop--可以

[YUNDAN21]

打卡

class Solution {
    public int[][] generateMatrix(int n) {
        int[][] res = new int[n][n];
        int k = 1;
        int t = 0;
        if (n % 2 == 1)
            res[n / 2][n / 2] = n * n;
        while(t < n / 2){
            for (int j = 0; j < 4; j++){
                for(int p = 0; p < n-1-2*t;p++){
                    if(j == 0)
                    {
                        res[t][p+t] = k;
                        k++;
                    }
                    else if(j == 1)
                    {
                        res[p+t][n-t-1] = k;
                        k++;
                    }
                    else if(j == 2)
                    {
                        res[n-t-1][n - p - 1 - t] = k;
                        k++;
                    }
                    else if(j==3)
                    {
                        res[n - p - 1 - t][t] = k;
                        k++;
                    }
                }
            }
            t += 1;
        }
        return res;
    }
}

[ltian059]

class Solution {
    public int[][] generateMatrix(int n) {
        int row = 0; // row pointer
        int col = 0; // column pointer
        int count = 1; // The actual number we are putting in the matrix
        int loopCount = 1;
        int startRow = 0; // The row number where we start to draw the matrix in the specifc loop.
        int startCol = 0;
        int[][] res = new int[n][n];

        int loop = n / 2; // The total loop number needed to draw the matrix.
        while(loopCount <= loop){
            col = startCol; // Make sure to start the column/row in the right start point of each loop;
            row = startRow; 

            /* Going from left column to the right column */
            for(; col < n - loopCount; col++){
                //row == 0, col's range :[0, n - loopCount)
                res[row][col] = count++;
            }

            /* Going from the upper row to the bottom row */
            for(; row < n - loopCount; row++){
                //row's range: [0, n - loopCount); col == n - loopCount
                res[row][col] = count++;
            }

            /* Going from the right column to the left column */
            for(; col > startCol; col--){
                //col's range: [n - loopCount, startCol); row == n - loopCount
                res[row][col] = count++;
            }

            /* Going from the bottom row to the upper row */
            for(; row > startRow; row--){
                //row's range: [n - loopCount, startRow); col == startCol
                res[row][col] = count++;
            }

            /* Update the start point of the next loop */
            startRow++;
            startCol++;

            loopCount++;
        }

        //When n is a odd number , the center point of the matrix cannot be reached in the loop, because we are using the [) interval.
        if((n % 2) != 0){
            res[n/2][n/2] = count;
        }

        return res;

    }
}

[yekater9420]

打卡 循环变量不变原则

[Ramodler]

1111

[dkddu]

打卡