【LeetCode】C# 59、Spiral Matrix II

4000
Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.

For example,

Given n = 3,

You should return the following matrix:

[

[ 1, 2, 3 ],

[ 8, 9, 4 ],

[ 7, 6, 5 ]

]

给定数字n返回一个n阶方阵，方阵中的数字按螺旋的形式从左上角的1排列到方阵中心。

思路，跟Spiral Matrix I 一样，可以建立rowBegin、End，colBegin、End来定义行列的起始位置，然后一圈一圈往中间遍历。

public static int[,] generateMatrix(int n) {
int[,] ret = new int[n,n];
int left = 0,top = 0;
int right = n -1,down = n - 1;
int count = 1;
while (left <= right) {
for (int j = left; j <= right; j ++) {
ret[top,j] = count++;
}
top ++;
for (int i = top; i <= down; i ++) {
ret[i,right] = count ++;
}
right --;
for (int j = right; j >= left; j --) {
ret[down,j] = count ++;
}
down --;
for (int i = down; i >= top; i --) {
ret[i,left] = count ++;
}
left ++;
}
return ret;
}

也可以定义一个计数器sum，然后根据第roundi圈和该圈没行列的关系列出循环。

public class Solution {
public int[,] GenerateMatrix(int n) {
int sum = 1;
int[,] res = new int[n,n];
for(int roundi=0;roundi<(n+1)/2;roundi++){
if(n-roundi-1==roundi){
res[roundi,roundi]=sum;
return res;
}
//top
for(int i=roundi;i<n-roundi-1;i++){
res[roundi,i] = sum++;
}
//right
for(int i=roundi;i<n-roundi-1;i++){
res[i,n-roundi-1]=sum++;
}
//bottom
for(int i=n-roundi-1;i>roundi;i--){
res[n-roundi-1,i] = sum++;
}
//left
for(int i=n-roundi-1;i>roundi;i--){
res[i,roundi] = sum++;
}
}
return res;
}
}