leetcode 54. Spiral Matrix 谷歌面试题
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.Example 1:
Input:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
Output:
Example 2:
Input:
[
,
,
]
Output:
思路：两个pointer，一个横着走，一个竖着走，每次计算matrix 的最外围一周，然后进行递归。考虑特殊情况，一个是empty set，另一个是size one array。
class Solution {
public static List<Integer> spiralOrder(int[][] matrix) {
ArrayList<Integer> recept = new ArrayList<>();
ArrayList<Integer> empty = new ArrayList<Integer>(){};
if(matrix == null || matrix.length == 0){
return empty;
}
if(matrix.length == 1){
for(int i = 0; i<matrix.length; i++ ){
recept.add(matrix);
}
return recept;
}
generateList(matrix, recept, matrix.length, matrix.length, 0);
return recept;
}
public static void generateList(int[][] matrix, ArrayList<Integer> array, int rowlen, int colen, int n) {
if(matrix == null){
return;
}
int len_of_column = colen - 2*n;
int len_of_row = rowlen - 2*n;
if(len_of_column <= 0 || len_of_row <= 0) {
return;
}
int first = 0 + n, second = 0 + n;
if(first == n) {
while(first != n+ len_of_row && first>=n && first <= n+ len_of_row) {
array.add(matrix[(first++)]);
// System.out.println(array);
// System.out.println("-----");
}
}
second++;
if(first == len_of_row+n && second == n+1) {
while(second != len_of_column + n && second >=n && second <= len_of_column + n) {
array.add(matrix);
// System.out.println(array);
// System.out.println("-----");
}
}
first= first -2 ;
if(first == len_of_row-2+n && second == len_of_column+n) {
while(first != 0+n && first >=n && first <= len_of_row+n) {
array.add(matrix[(first--)]);
// System.out.println(array);
// System.out.println("-----");
}
}
second--;
if(first == 0+n && second == len_of_column-1+n) {
while(second != n && second >=n && second <=len_of_column+n) {
array.add(matrix[(second--)]);
}
}
n++;
generateList(matrix, array, rowlen, colen, n);
}
}
结果在running time 和 memory usage beat 100% online submissions！！！
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