Matrix Zigzag Traversal
最后更新于：2022-04-02 01:14:18
# Matrix Zigzag Traversal
### Source
- lintcode: [(185) Matrix Zigzag Traversal](http://www.lintcode.com/en/problem/matrix-zigzag-traversal/)
~~~
Given a matrix of m x n elements (m rows, n columns),
return all elements of the matrix in ZigZag-order.
Example
Given a matrix:
[
[1, 2, 3, 4],
[5, 6, 7, 8],
[9,10, 11, 12]
]
return [1, 2, 5, 9, 6, 3, 4, 7, 10, 11, 8, 12]
~~~
### 题解
按之字形遍历矩阵，纯粹找下标规律。以题中所给范例为例，设`(x, y)`为矩阵坐标，按之字形遍历有如下规律：
~~~
(0, 0)
(0, 1), (1, 0)
(2, 0), (1, 1), (0, 2)
(0, 3), (1, 2), (2, 1)
(2, 2), (1, 3)
(2, 3)
~~~
可以发现其中每一行的坐标之和为常数，坐标和为奇数时 x 递增，为偶数时 x 递减。
### Java - valid matrix index second
~~~
public class Solution {
/**
* @param matrix: a matrix of integers
* @return: an array of integers
*/
public int[] printZMatrix(int[][] matrix) {
if (matrix == null || matrix.length == 0) return null;
int m = matrix.length - 1, n = matrix[0].length - 1;
int[] result = new int[(m + 1) * (n + 1)];
int index = 0;
for (int i = 0; i <= m + n; i++) {
if (i % 2 == 0) {
for (int x = i; x >= 0; x--) {
// valid matrix index
if ((x <= m) && (i - x <= n)) {
result[index] = matrix[x][i - x];
index++;
}
}
} else {
for (int x = 0; x <= i; x++) {
if ((x <= m) && (i - x <= n)) {
result[index] = matrix[x][i - x];
index++;
}
}
}
}
return result;
}
}
~~~
### Java - valid matrix index first
~~~
public class Solution {
/**
* @param matrix: a matrix of integers
* @return: an array of integers
*/
public int[] printZMatrix(int[][] matrix) {
if (matrix == null || matrix.length == 0) return null;
int m = matrix.length - 1, n = matrix[0].length - 1;
int[] result = new int[(m + 1) * (n + 1)];
int index = 0;
for (int i = 0; i <= m + n; i++) {
int upperBoundx = Math.min(i, m); // x <= m
int lowerBoundx = Math.max(0, i - n); // lower bound i - x(y) <= n
int upperBoundy = Math.min(i, n); // y <= n
int lowerBoundy = Math.max(0, i - m); // i - y(x) <= m
if (i % 2 == 0) {
// column increment
for (int y = lowerBoundy; y <= upperBoundy; y++) {
result[index] = matrix[i - y][y];
index++;
}
} else {
// row increment
for (int x = lowerBoundx; x <= upperBoundx; x++) {
result[index] = matrix[x][i - x];
index++;
}
}
}
return result;
}
}
~~~
### 源码分析
矩阵行列和分奇偶讨论，奇数时行递增，偶数时列递增，一种是先循环再判断索引是否合法，另一种是先取的索引边界。
### 复杂度分析
后判断索引是否合法的实现遍历次数为 1+2+...+(m+n)=O((m+n)2)1 + 2 + ... + (m + n) = O((m+n)^2)1+2+...+(m+n)=O((m+n)2), 首先确定上下界的每个元素遍历一次，时间复杂度 O(m⋅n)O(m \cdot n)O(m⋅n). 空间复杂度都是 O(1)O(1)O(1).
### Reference
- [LintCode/matrix-zigzag-traversal.cpp at master · kamyu104/LintCode](https://github.com/kamyu104/LintCode/blob/master/C++/matrix-zigzag-traversal.cpp)
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