剑指offer (20) 打印螺旋矩阵

题目描述：

输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数字

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int> > &matrix) {
        vector<int> result;
        int nRows = matrix.size();
        if (nRows == 0) return result;
        int nCols = matrix.at(0).size();
        result.reserve(nRows * nCols);

        int startX = 0;
        int startY = 0;

        while (nRows > 2 * startX  && nCols > 2 * startY) {
            PrintNumbers(matrix, startX, startY, result);
            ++startX;
            ++startY;
        }
        return result;

    }
        void PrintNumbers(vector<vector<int>>& matrix, int startX, int startY, vector<int>& result) {
            assert(startX >= 0 && startY >= 0);
            int nRows = matrix.size();
            int nCols = matrix.at(0).size();

            int rowLen = nRows - 2 * startX; // 当前矩形的长度
            int colLen = nCols - 2 * startY; // 当前矩形的宽度

            int endX = startX + colLen - 1;  // endX - startX + 1 = colLen X为横坐标
            int endY = startY + rowLen - 1;  // endY - startY + 1 = rowLen Y为纵坐标

            // 1. 从左往右
            for (int i = startX; i <= endX; ++i) {
                result.push_back(matrix.at(startY).at(i));
            }

            // 2. 从上往下
            if (endY > startY) {
                for (int i = startY + 1; i <= endY; ++i) {
                    result.push_back(matrix.at(i).at(endX));
                }
            }

            // 3. 从右往左
            if (endX > startX && endY > startY) {
                for (int i = endX - 1; i >= startX; --i) {
                    result.push_back(matrix.at(endY).at(i));
                }
            }

            // 4. 从下往上
            if (endX > startX && endY > startY) {
                for (int i = endY - 1; i >= startY + 1; --i) {
                    result.push_back(matrix.at(i).at(startX));
                }
            }
        }
};

具体分析详见我的博文：螺旋矩阵

剑指offer (20) 打印螺旋矩阵