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]
public List<Integer> spiralOrder(int[][] matrix) {
int left = 0, right = matrix[0].length - 1;
int top = 0, bottom = matrix.length - 1;
List<Integer> res = new ArrayList<>();
int size = matrix.length * matrix[0].length;
while (true) {
for (int i = left; i <= right; i++) {
res.add(matrix[top][i]);
}
top++;
if (res.size() == size) {
break;
}
for (int i = top; i <= bottom; i++) {
res.add(matrix[i][right]);
}
right--;
if (res.size() == size) {
break;
}
for (int i = right; i >= left; i--) {
res.add(matrix[bottom][i]);
}
bottom--;
if (res.size() == size) {
break;
}
for (int i = bottom; i >= top; i--) {
res.add(matrix[i][left]);
}
left++;
if (res.size() == size) {
break;
}
}
return res;
}
螺旋矩阵 II
给你一个正整数 n
,生成一个包含 1
到 n<sup>2</sup>
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]]
public int[][] generateMatrix(int n) {
int all = n * n;
int[][] res = new int[n][n];
int k = 0;
int left = 0, right = n - 1, top = 0, bottom = n - 1;
while (left <= right && top <= bottom) {
if (k < all) {
for (int i = left; i <= right; i++) {
res[top][i] = ++k;
}
top++;
}
if (k < all) {
for (int i = top; i <= bottom; i++) {
res[i][right] = ++k;
}
right--;
}
if (k < all) {
for (int i = right; i >= left; i--) {
res[bottom][i] = ++k;
}
bottom--;
}
if (k < all) {
for (int i = bottom; i >= top; i--) {
res[i][left] = ++k;
}
left++;
}
}
return res;
}