59. Spiral Matrix II（待进一步研究）

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 ]

]

四步走：

vector<vector<int>> generateMatrix(int n) {
vector<vector<int>> result(n,vector<int>(n));
int k = 1;
int i = 0;
while(k <= n * n){
//四个步骤
int j = i;
while(j < n - i){ //水平赋值，从左到右
result[i][j++] = k++;
}

j = i + 1;
while(j < n - i){//垂直赋值，从上到下
result[j++][n - i - 1] = k++;
}

j = n - i - 2;
while(j > i){//水平赋值，从右到左
result[n - i - 1][j--] = k++;
}

j = n - i -1;
while(j > i){//垂直赋值，从下到上
result[j--][i] = k++;
}

i++;
}

return result;
}

也可以采用定义的4个变量来写,这样更清晰些（注意：此题中注释掉的两个if语句可加可不加，加上更严谨些，但注意对于n*n的矩阵可不加if,但对于m*n的矩阵则需要加上if）：

vector<vector<int>> generateMatrix(int n) {
vector<vector<int>> result(n,vector<int>(n));
int rowStart = 0;
int rowEnd = n - 1;
int colStart = 0;
int colEnd = n -1;
int k = 1;

while(rowStart <= rowEnd && colStart <= colEnd){
//四步走
//第一步：水平赋值，从左到右
for(int i = colStart; i <= colEnd; i++)
{
result[rowStart][i] = k++;
}
rowStart++;

//第二步，垂直赋值，从上到下
for(int i = rowStart; i <= rowEnd; i++)
{
result[i][colEnd] = k++;
}
colEnd--;

//第三步，水平赋值，从右到左
for(int i = colEnd; i >= colStart; i--)
{
//if (rowStart <= rowEnd)
result[rowEnd][i] = k++;
}
rowEnd--;

//第四步,垂直赋值，从下到上
for(int i = rowEnd; i >= rowStart; i--)
{
//if (colStart <= colEnd)
result[i][colStart] = k++;
}
colStart++;

}
return result;
}

更好的算法：

vector<vector<int> > generateMatrix(int n) {
int dir = 0;
vector< vector<int> > matrix(n, vector<int> (n, 0));
int i = 0, j = 0, k = 1;
while (k <= n * n) {
matrix[i][j] = k++;
if (dir == 0){
j++;
if (j == n || matrix[i][j] != 0) dir = 1, j--, i++;
} else
if (dir == 1) {
i++;
if (i == n || matrix[i][j] != 0) dir = 2, i--, j--;
} else
if (dir == 2) {
j--;
if (j < 0 || matrix[i][j] != 0) dir = 3, j++, i--;
} else
if (dir == 3) {
i--;
if (i < 0 || matrix[i][j] != 0) dir = 0, i++, j++;
}
}
return matrix;
}