題目地址:
https://leetcode.com/problems/diagonal-traverse/
給定一個二維矩陣行列,顯然形如從左下到右上的對角線,一共有條。現在從右上到左下將這些對角線編號爲,要求按照編號爲偶數的對角線從左下到右上遍歷,編號爲奇數的對角線從右上到左下遍歷。返回遍歷的結果。
可以先遍歷矩陣左上半部分的對角線,然後遍歷右下半部分的對角線,並且記錄遍歷的順序是順序還是逆序。代碼如下:
import java.util.ArrayList;
import java.util.List;
public class Solution {
public int[] findDiagonalOrder(int[][] matrix) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return new int[0];
}
int n = matrix.length, m = matrix[0].length;
int[] res = new int[n * m];
int idx = 0;
boolean reversed = true;
for (int i = 0; i < m; i++) {
List<Integer> diag = new ArrayList<>();
int x = 0, y = i;
while (x < n && y >= 0) {
diag.add(matrix[x][y]);
x++;
y--;
}
if (!reversed) {
for (int j = 0; j < diag.size(); j++) {
res[idx++] = diag.get(j);
}
} else {
for (int j = diag.size() - 1; j >= 0; j--) {
res[idx++] = diag.get(j);
}
}
// 遍歷完一條對角線後改變reversed的值
reversed = !reversed;
}
for (int i = 1; i < n; i++) {
List<Integer> diag = new ArrayList<>();
int x = i, y = m - 1;
while (x < n && y >= 0) {
diag.add(matrix[x][y]);
x++;
y--;
}
if (!reversed) {
for (int j = 0; j < diag.size(); j++) {
res[idx++] = diag.get(j);
}
} else {
for (int j = diag.size() - 1; j >= 0; j--) {
res[idx++] = diag.get(j);
}
}
reversed = !reversed;
}
return res;
}
}
時間複雜度,空間。