力扣-数组-59 螺旋矩阵Ⅱ

思路1

记录当前前进方向，然后在到达上下左右的边界时，分别进行处理

代码1

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        vector< vector<int> > res;
        for(int i = 0; i < n; i++){
            vector <int> r;
            for(int j = 0; j < n; j++){
                r.push_back(0);
            }
            res.push_back(r);
        }

        int num = 1;
        int flag = 0;// 0 right 1 down 2 left 3 up
        int right = n - 1, down = n - 1, left = 0, up = 1;
        int i=0, j = 0;    

        while(num <= n*n){
            res[i][j] = num;
            if(flag == 0){
                if(j < right){
                    j++;
                }else{
                    flag = (flag + 1) % 4;
                    right -= 1;
                    i++;
                }
            }else if(flag == 1){
                if(i < down){
                    i++;
                }else{
                    flag = (flag + 1) % 4;
                    down -= 1;
                    j--;
                }
            }else if(flag == 2){
                if(j > left){
                    j--;
                }else{
                    flag = (flag + 1) % 4;
                    left += 1;
                    i--;
                }
            }else if(flag == 3){
                if(i > up){
                    i--;
                }else{
                    flag = (flag + 1) % 4;
                    up += 1;
                    j++;
                }
            }


            num += 1;
        }

        return res;
                
    }
};

思路2

根据n分解为n/2圈，然后在一圈中用四个for循环更新边，最后更改下一圈的起始位置

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        vector< vector<int> > res;
        for(int i = 0; i < n; i++){
            vector <int> r;
            for(int j = 0; j < n; j++){
                r.push_back(0);
            }
            res.push_back(r);
        }

        int num = 1, count = n/2, startX = 0, startY = 0;
        int i = 0, j = 0;
        int offset = 1;
        while(count){
            for(j = startY; j < n - offset; j++){
                res[startX][j] = num;
                num++;
            }
            for(i = startX; i < n - offset; i++){
                res[i][j] = num;
                num++;
            }
            for(; j > startY; j--){
                res[i][j] = num;
                num++;
            }
            for(; i > startX; i--){
                res[i][j] = num;
                num++;
            }
            offset++;
            count--;
            startX += 1;
            startY += 1;
        }

        if(n % 2 == 1){
            res[startX][startY] = num;
        }

        return res;
                
    }
};