498. Diagonal Traverse
2017-03-10 22:23
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Given a matrix of M x N elements (M rows, N columns), return all elements of the matrix in diagonal order as shown in the below image.
Example:
Note:
The total number of elements of the given matrix will not exceed 10,000.
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My solution:
public class Solution {
public int[] findDiagonalOrder(int[][] matrix) {
if (matrix == null || matrix.length == 0) return new int[0];
int m = matrix.length;
int n = matrix[0].length;
int[] re = new int[m * n];
boolean flag = true;
int mnow = 0;
int nnow = 0;
for (int i = 0; i < m * n; ++i) {
re[i] = matrix[mnow][nnow];
if (flag) {
if (mnow == 0 || nnow == n - 1) {
if (nnow == n - 1)
mnow++;
else
nnow++;
flag = false;
} else {
mnow--;
nnow++;
}
} else {
if (mnow == m - 1 || nnow == 0) {
if (mnow == m - 1)
nnow++;
else
mnow++;
flag = true;
} else {
mnow++;
nnow--;
}
}
}
return re;
}
}
4000
Example:
Input: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] Output: [1,2,4,7,5,3,6,8,9] Explanation:
Note:
The total number of elements of the given matrix will not exceed 10,000.
Subscribe to see which companies asked this question.
My solution:
public class Solution {
public int[] findDiagonalOrder(int[][] matrix) {
if (matrix == null || matrix.length == 0) return new int[0];
int m = matrix.length;
int n = matrix[0].length;
int[] re = new int[m * n];
boolean flag = true;
int mnow = 0;
int nnow = 0;
for (int i = 0; i < m * n; ++i) {
re[i] = matrix[mnow][nnow];
if (flag) {
if (mnow == 0 || nnow == n - 1) {
if (nnow == n - 1)
mnow++;
else
nnow++;
flag = false;
} else {
mnow--;
nnow++;
}
} else {
if (mnow == m - 1 || nnow == 0) {
if (mnow == m - 1)
nnow++;
else
mnow++;
flag = true;
} else {
mnow++;
nnow--;
}
}
}
return re;
}
}
4000
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