【Lee Button】59. Spiral Matrix II
give you a positive integer nnn , generate a list containing 1 ton 2 n^2n2 All elements, and the elements are spirally arranged in clockwise ordern × nn × nn×n square matrixmatrix matrixmatrix 。
Example 1:
Input: n=3
Output: [[1,2,3],[8,9,4],[7,6,5]]
Example 2:
Input: n = 1
Output: [[1]]
Hint:
1 <= n <= 20
answer
Pay attention to the principle of loop invariance during each loop
- Upstream from left to right (left closed right open)
- Right column from top to bottom (left closed right open)
- Downstream from right to left (left closed right open)
- Left column from bottom to top (left closed right open)
public class Main {
public int[][] generateMatrix(int n) {
int[][] res = new int[n][n];
// 定义每循环一个圈的起始位置
int startx = 0, starty = 0;
// 每个圈循环几次,例如 n为奇数 3,那么loop = 1 只是循环一圈,矩阵中间的值需要单独处理
int loop = n / 2;
// 矩阵中间的位置,例如:n为 3, 中间的位置就是(1,1),n为5,中间位置为(2, 2)
int mid = n / 2;
// 用来计数赋值
int count = 1;
// 需要控制每一条边遍历的长度,每次循环右边界收缩一位
int offset = 1;
int i, j;
while ((loop--) > 0) {
i = startx;
j = starty;
// 模拟填充上行从左到右(左闭右开)
for (j = starty; j < n - offset; j++) {
res[i][j] = count++;
}
// 模拟填充右列从上到下(左闭右开)
for (i = startx; i < n - offset; i++) {
res[i][j] = count++;
}
// 模拟填充下行从右到左(左闭右开)
for (; j > starty; j--) {
res[i][j] = count++;
}
// 模拟填充左列从下到上(左闭右开)
for (; i > startx; i--) {
res[i][j] = count++;
}
// 第二圈开始的时候,起始位置要各自加1, 例如:第一圈起始位置是(0, 0),第二圈起始位置是(1, 1)
startx++;
starty++;
// offset 控制每一圈里每一条边遍历的长度
offset += 1;
}
// 如果n为奇数的话,需要单独给矩阵最中间的位置赋值
if ((n % 2) == 1) {
res[mid][mid] = count; //res[startx][starty] = count;
}
return res;
}
}