原文引用https://www.dazhuanlan.com/2019/08/26/5d62fafb4de84/
题目大意
https://leetcode.com/problems/spiral-matrix/
https://leetcode.com/problems/spiral-matrix-ii/
这两个题目本质上是一道题，都是螺旋式或者生成打印矩阵
题目分析
最开始我的实现还是每行或者每列少打印一个，留给下一列或者下一行处理，这样导致最后还要特殊判断到底是剩下1行/列，或者是两行/列。导致代码极其不优雅，虽然可以AC。我觉得刷leetcode目标不仅仅是AC，更要追求时间/空间复杂度，以及代码的优雅。看了disscuss区的高票答案，就是简单的每次“缩减”行或者列就可以很优雅的解出来。至于Spiral Matrix II反而更简单，题目要求的是生成螺旋正方形，可以几乎完全复用代码即可。
代码
Spiral Matrix:
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public
class {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> ans =
new ArrayList<Integer>();
if (matrix ==
null || matrix.length ==
0 || matrix[
0].length ==
0) {
return ans;
}
int m = matrix.length;
int n = matrix[
0].length;
int up =
0;
int down = m -
1;
int left =
0;
int right = n -
1;
while (
true) {
for (
int col = left; col <= right; col++) {
ans.add(matrix[up][col]);
}
up++;
if (up > down) {
break;
}
for (
int row = up; row <= down; row++) {
ans.add(matrix[row][right]);
}
right--;
if (right < left) {
break;
}
for (
int col = right; col >= left; col--) {
ans.add(matrix[down][col]);
}
down--;
if (down < up) {
break;
}
for (
int row = down; row >= up; row--) {
ans.add(matrix[row][left]);
}
left++;
if (left > right) {
break;
}
}
return ans;
}
}
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Spiral Matrix II:
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public
class Solution {
public
int[][] generateMatrix(
int n) {
int[][] matrix =
new
int[n][n];
int up =
0;
int down = n -
1;
int left =
0;
int right = n -
1;
int c =
0;
while (
true) {
for (
int col = left; col <= right; col++) {
matrix[up][col] = ++c;
}
up++;
if (up > down) {
break;
}
for (
int row = up; row <= down; row++) {
matrix[row][right] = ++c;
}
right--;
if (right < left) {
break;
}
for (
int col = right; col >= left; col--) {
matrix[down][col] = ++c;
}
down--;
if (down < up) {
break;
}
for (
int row = down; row >= up; row--) {
matrix[row][left] = ++c;
}
left++;
if (left > right) {
break;
}
}
return matrix;
}
}
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