leetcode_spiral_order

发表于

难度：Medium

解题思路：按照顺时针一圈一圈的取值，注意每一圈的边界。
代码如下。

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class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector<int> ret;
        if(matrix.empty()) return ret;
        size_t current_row_num = matrix.size();
        size_t current_col_num = matrix[0].size();
        int left_top_bound = 0;
        while(current_row_num != 0 && current_col_num != 0)
        {
            int i = left_top_bound ,j = left_top_bound;
            for(; j < current_col_num+left_top_bound ; j++)
            {
                ret.push_back(matrix[i][j]);
            }
            j--;
            i++;
            for(; i < current_row_num+left_top_bound; i++)
            {
                ret.push_back(matrix[i][j]);
            }
            i--;
            j--;
            for(; j >= left_top_bound && current_row_num > 1; j--)
            {
                ret.push_back(matrix[i][j]);
            }
            j++;
            i--;
            for(; i > left_top_bound && current_col_num > 1; i--)
            {
                ret.push_back(matrix[i][j]);
            }
            i++;
            left_top_bound++;
            current_row_num = (current_row_num >= 2)?current_row_num-2:current_row_num-1;
            current_col_num = (current_col_num >= 2)?current_col_num-2:current_col_num-1;
        }
        return ret;
    }
};

运行结果：0ms(0~2.5ms区间)，超过26.22%