29.顺时针打印矩阵 (Easy)*

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。

示例*

输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]

输出：[1,2,3,6,9,8,7,4,5]

输入：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]

输出：[1,2,3,4,8,12,11,10,9,5,6,7]

限制*

0 <= matrix.length <= 100, 0 <= matrix[i].length <= 100

代码*

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector<int> res;
        if(matrix.size() == 0 || matrix[0].size() == 0) {
            return res;
        }
        int targetLength = matrix.size() * matrix[0].size();
        int rowTop = 0, rowBottom = matrix.size() - 1, colLeft = 0, colRight = matrix[0].size() - 1;
        while(rowTop <= rowBottom && colLeft <= colRight) {
            for(int i = colLeft; i <= colRight; i++) {
                res.push_back(matrix[rowTop][i]);
            }
            for(int i = rowTop + 1; i <= rowBottom; i++) {
                res.push_back(matrix[i][colRight]);
            }
            if(rowBottom != rowTop) {
                for(int i = colRight - 1; i >= colLeft; i--) {
                    res.push_back(matrix[rowBottom][i]);
                }
            }
            if(colLeft != colRight) {
                for(int i = rowBottom - 1; i > rowTop; i--) {
                    res.push_back(matrix[i][colLeft]);
                }
            }
            colLeft++, colRight--, rowTop++, rowBottom--;
        }
        return res;
    }
};

最后更新: July 23, 2022