[C语言][LeetCode][59]Spiral Matrix II

题目

Spiral Matrix II

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 ]

]

标签

Array

难度

中等

分析

题目意思是给定一个整形n，创建一个旋转的矩阵，由1-n²的数字填充。

解题思路是，按照以下顺序遍历数组，直到循环结束

1. 从colBegin到colEnd遍历，rowBegin++

2. 从rowBegin到rowEnd遍历，colEnd–

3. 从colEnd到colBegin遍历，rowEnd–

4. 从rowEnd到rowBegin遍历，colBegin++

C代码实现

int** generateMatrix(int n) {
int i=0, num=1;
int rowBegin=0, rowEnd= n-1;
int colBegin=0, colEnd=n-1;

int **matrix= (int **)malloc(n*sizeof(int *));
for(i=0;i<n;i++)
{
matrix[i]=(int *)malloc(n*sizeof(int));
memset(matrix[i], 0, sizeof(int)*n);
}

while(rowBegin<=rowEnd && colBegin<=colEnd)
{
for(i=colBegin; i<=colEnd; i++)
{
matrix[rowBegin][i] = num++;
//printf("111 matrix[%d][%d] : %d\n", rowBegin, i, matrix[rowBegin][i]);
}
rowBegin++;

for(i=rowBegin; i<=rowEnd; i++)
{
matrix[i][colEnd] = num++;
//printf("222 matrix[%d][%d] : %d\n", i, colEnd, matrix[i][colEnd]);
}
colEnd--;

if(rowBegin<=rowEnd)
{
for(i=colEnd; i>=colBegin; i--)
{
matrix[rowEnd][i] = num++;
//printf("333 matrix[%d][%d] : %d\n", rowEnd, i, matrix[rowEnd][i]);
}
}
rowEnd--;

if(colBegin<=colEnd)
{
for(i=rowEnd; i>=rowBegin; i--)
{
matrix[i][colBegin] = num++;
//printf("444 matrix[%d][%d] : %d\n", i, colBegin, matrix[i][colBegin]);
}
}
colBegin++;
}

return matrix;
}