[leetcode_54]Spiral Matrix

螺旋输出矩阵 注意一行和一列的边界情况

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class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>> &matrix) {
        vector<int> ans;
        ans.clear();
        if(matrix.size() <= 0) return ans;
        int count = 0;
        int sum = matrix.size() * matrix[0].size();
        int px = 0;
        int py = 0;

        int xa = 0;
        int xb = matrix.size() - 1;
        int ya = 0;
        int yb = matrix[0].size() - 1;

        int dir = 0;
        while(count < sum) {
            ans.push_back(matrix[px][py]);
            switch(dir) {
                case 0: 
                    if(py == yb) {
                        dir++;
                        dir %= 4;
                        xa++;
                        px++;
                        break;
                    }
                    py++;
                    if(py == yb) {
                        dir++;
                        dir %= 4;
                        xa++;
                    }
                    break;
                case 1:
                    if(px == xb) {
                        dir++;
                        dir %= 4;
                        yb--;
                        py--;
                        break;
                    }
                    px++;
                    if(px == xb) {
                        dir++;
                        dir %= 4;
                        yb--;
                    }
                    break;  
                case 2:
                    if(py == ya) {
                        dir++;
                        dir %= 4;
                        xb--;
                        px--;
                        break;
                    }
                    py--;
                    if(py == ya) {
                        dir++;
                        dir %= 4;
                        xb--;
                    }
                    break;
                case 3:
                    if(px == xa) {
                        dir++;
                        dir %= 4;
                        ya++;
                        py++;
                        break;
                    }
                    px--;
                    if(px == xa) {
                        dir++;
                        dir %= 4;
                        ya++;
                    }
                    break;
            }
            count++;
        }
        return ans;
    }
};
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