59. Spiral Matrix II (Array)
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 ]
]
class Solution {
public:
vector<vector<int> > generateMatrix(int n) {
vector<vector<int> > result(n, vector<int>(n,));
leftPos = ;
rightPos = n-;
topPos = ;
bottomPos = n-;
currentNum = ;
goWider(result,true);
return result;
}
void goWider(vector<vector<int> > &matrix, bool direct)
{
if(direct)
{
for(int i = leftPos; i<= rightPos; i++)
{
matrix[topPos][i] = currentNum++;
}
topPos++;
if(topPos > bottomPos) return;
goDeeper(matrix, true);
}
else
{
for(int i = rightPos; i>= leftPos; i--)
{
matrix[bottomPos][i] = currentNum++;
}
bottomPos--;
if(topPos > bottomPos) return;
goDeeper(matrix, false);
}
}
void goDeeper(vector<vector<int> > &matrix, bool direct)
{
if(direct)
{
for(int i = topPos; i<= bottomPos; i++)
{
matrix[i][rightPos]=currentNum++;
}
rightPos--;
if(leftPos > rightPos) return;
goWider(matrix, false);
}
else
{
for(int i = bottomPos; i>= topPos; i--)
{
matrix[i][leftPos] = currentNum++;
}
leftPos++;
if(leftPos > rightPos) return;
goWider(matrix, true);
}
}
private:
int currentNum;
int leftPos;
int rightPos;
int topPos;
int bottomPos;
};
思路II:向右向下向左向上,4个for循环。外套一个while,while的结束条件是4个for循环都不满足循环条件了。
class Solution {
public:
vector<vector<int>> generateMatrix(int n) {
int len = n;
int num = ;
vector<vector<int>> ret(n, vector<int>(n,));
while(len>n-len){
for(int i = n-len; i < len; i++){
ret[n-len][i] = num++;
}
for(int j = n-len+; j < len; j++){
ret[j][len-] = num++;
}
for(int k = len-; k >= n-len; k--){
ret[len-][k] = num++;
}
for(int l = len-; l > n-len; l--){
ret[l][n-len] = num++;
}
len--;
}
return ret;
}
};
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