54. Spiral Matrix
题目:
Given a matrix of m x n elements (m rows, ncolumns), return all elements of the matrix in spiral order.
For example,
Given the following matrix:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
You should return [1,2,3,6,9,8,7,4,5]
.
链接: http://leetcode.com/problems/spiral-matrix/
题解:
转圈打印矩阵,需要设置left, right, top和bot四个变量来控制, 注意每次增加行/列前要判断list所含元素是否小于矩阵的总元素数目。需要再想办法简化一下。
Time Complexity - O(mn), Space Complexity - O(1)
public class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> res = new ArrayList<>();
if(matrix == null || matrix.length == 0)
return res;
int rowNum = matrix.length, colNum = matrix[0].length;
int left = 0, right = colNum - 1, top = 0, bot = rowNum - 1; while(res.size() < rowNum * colNum) {
for(int col = left; col <= right; col++)
res.add(matrix[top][col]);
top++;
if(res.size() < rowNum * colNum) {
for(int row = top; row <= bot; row++)
res.add(matrix[row][right]);
right--;
}
if(res.size() < rowNum * colNum) {
for(int col = right; col >= left; col--)
res.add(matrix[bot][col]);
bot--;
}
if(res.size() < rowNum * colNum) {
for(int row = bot; row >= top; row--)
res.add(matrix[row][left]);
left++;
}
} return res;
}
}
二刷:
跟一刷一样,设置上下左右四边界,然后编写就可以了。
Java:
Time Complexity - O(mn), Space Complexity - O(1)
public class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> res = new ArrayList<>();
if (matrix == null || matrix.length == 0) {
return res;
}
int rowNum = matrix.length, colNum = matrix[0].length;
int top = 0, bot = matrix.length - 1, left = 0, right = colNum - 1;
int elementsLeft = rowNum * colNum;
while (elementsLeft >= 0) {
if (elementsLeft >= 0) {
for (int i = left; i <= right; i++) {
res.add(matrix[top][i]);
elementsLeft--;
}
top++;
}
if (elementsLeft >= 0) {
for (int i = top; i <= bot; i++) {
res.add(matrix[i][right]);
elementsLeft--;
}
right--;
}
if (elementsLeft >= 0) {
for (int i = right; i >= left; i--) {
res.add(matrix[bot][i]);
elementsLeft--;
}
bot--;
}
if (elementsLeft >= 0) {
for (int i = bot; i >= top; i--) {
res.add(matrix[i][left]);
elementsLeft--;
}
left++;
}
}
return res;
}
}
三刷:
条件是total > 0就可以了,不需要">="。
Java:
public class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> res = new ArrayList<>();
if (matrix == null || matrix.length == 0) return res;
int rowNum = matrix.length, colNum = matrix[0].length;
int left = 0, top = 0, right = colNum - 1, bot = rowNum - 1;
int total = rowNum * colNum;
while (total > 0) {
for (int i = left; i <= right; i++) {
res.add(matrix[top][i]);
total--;
}
top++; if (total > 0) {
for (int i = top; i <= bot; i++) {
res.add(matrix[i][right]);
total--;
}
right--;
} if (total > 0) {
for (int i = right; i >= left; i--) {
res.add(matrix[bot][i]);
total--;
}
bot--;
} if (total > 0) {
for (int i = bot; i >= top; i--) {
res.add(matrix[i][left]);
total--;
}
left++;
}
}
return res;
}
}
Reference:
https://leetcode.com/discuss/12228/super-simple-and-easy-to-understand-solution
https://leetcode.com/discuss/62196/ac-python-32ms-solution
https://leetcode.com/discuss/46523/1-liner-in-python
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