LeetCode Spiral Matrix II
2014-11-05 21:19
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Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.
For example,
Given n =
You should return the following matrix:
与上一题相反,对矩阵赋值,步骤相同。
<pre name="code" class="java">public class Solution {
public int[][] generateMatrix(int n) {
int[][] result=new int
;
int rowstart = 0, rowend = n, colstart = 0, colend = n;
int num = 0;
while (num != n * n) {
// 上行
for (int i = colstart; i < colend; i++) {
result[rowstart][i]=num+1;
num++;
if (num == n * n)
return result;
}
rowstart++;
// 右列
for (int i = rowstart; i < rowend; i++) {
result[i][colend - 1]=num+1;
num++;
if (num == n * n)
return result;
}
colend--;
// 下行
for (int i = colend - 1; i >= colstart; i--) {
result[rowend - 1][i]=num+1;
num++;
if (num == n * n)
return result;
}
rowend--;
// 左列
for (int i = rowend - 1; i >= rowstart; i--) {
result[i][colstart]=num+1;
num++;
if (num == n * n)
return result;
}
colstart++;
}
return result;
}
}
For example,
Given n =
3,
You should return the following matrix:
[ [ 1, 2, 3 ], [ 8, 9, 4 ], [ 7, 6, 5 ] ]
与上一题相反,对矩阵赋值,步骤相同。
<pre name="code" class="java">public class Solution {
public int[][] generateMatrix(int n) {
int[][] result=new int
;
int rowstart = 0, rowend = n, colstart = 0, colend = n;
int num = 0;
while (num != n * n) {
// 上行
for (int i = colstart; i < colend; i++) {
result[rowstart][i]=num+1;
num++;
if (num == n * n)
return result;
}
rowstart++;
// 右列
for (int i = rowstart; i < rowend; i++) {
result[i][colend - 1]=num+1;
num++;
if (num == n * n)
return result;
}
colend--;
// 下行
for (int i = colend - 1; i >= colstart; i--) {
result[rowend - 1][i]=num+1;
num++;
if (num == n * n)
return result;
}
rowend--;
// 左列
for (int i = rowend - 1; i >= rowstart; i--) {
result[i][colstart]=num+1;
num++;
if (num == n * n)
return result;
}
colstart++;
}
return result;
}
}
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