lintcode-185-矩阵的之字型遍历

185-矩阵的之字型遍历

给你一个包含 m x n 个元素的矩阵 (m 行, n 列), 求该矩阵的之字型遍历。

样例

对于如下矩阵：
[
[1, 2, 3, 4],
[5, 6, 7, 8],
[9,10, 11, 12]
]
返回 [1, 2, 5, 9, 6, 3, 4, 7, 10, 11, 8, 12]

标签

LintCode 版权所有 矩阵

思路

参考http://blog.csdn.net/wutingyehe/article/details/46629087

code

class Solution {
public:
    /**
     * @param matrix: a matrix of integers
     * @return: a vector of integers
     */
    vector<int> printZMatrix(vector<vector<int> > &matrix) {
        // write your code here
        int sizeRow = matrix.size();
        if (sizeRow <= 0) {
            return vector<int>();
        }
        int sizeCol = matrix[0].size();
        if (sizeCol <= 0) {
            return vector<int>();
        }

        vector<int> result;
        int count = sizeRow * sizeCol, curRow = 0, curCol = 0;
        result.push_back(matrix[0][0]);
        for (int i = 1; i < count; ) {
            //斜上走到顶
            while (i < count && curRow - 1 >= 0 && curCol + 1 < sizeCol) {
                result.push_back(matrix[--curRow][++curCol]);
                i++;
            }
            //横右走一步，不可横右走时竖下走一步
            if (i < count && curCol + 1 < sizeCol) {
                result.push_back(matrix[curRow][++curCol]);
                i++;
            }
            else if (i < count && curRow + 1 < sizeRow) {
                result.push_back(matrix[++curRow][curCol]);
                i++;
            }
            //斜下走到底
            while (i < count && curRow + 1 < sizeRow && curCol - 1 >= 0) {
                result.push_back(matrix[++curRow][--curCol]);
                i++;
            }
            //竖下走一步，不可竖下走时横右走一步
            if (i < count && curRow + 1 < sizeRow) {
                result.push_back(matrix[++curRow][curCol]);
                i++;
            }
            else if (i < count && curCol + 1 < sizeCol) {
                result.push_back(matrix[curRow][++curCol]);
                i++;
            }
        }
        return result;
    }
};