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 1:
输入:matrix = [[1,2,3],[4,5,6],[7,8,9]]
输出:[1,2,3,6,9,8,7,4,5]
从外往里一圈一圈遍历并存储矩阵元素即可。
class Solution { public: vector spiralOrder(vector>& matrix) { int rows = matrix.size(), cols = matrix[0].size(); if (rows == 0 || cols == 0) return {}; vector res; int top = 0, bottom = rows -1, left = 0, right = cols - 1; while (top <= bottom && left <= right) { for (int i = left; i <= right; i++) res.push_back(matrix[top][i]); for (int i = top + 1; i <= bottom; i++) res.push_back(matrix[i][right]); // 判断不可少 if (left < right && top < bottom) { for (int i = right - 1; i >= left; i--) res.push_back(matrix[bottom][i]); // 可以模拟发现 不可以为 i >= top for (int i = bottom - 1; i > top; i--) res.push_back(matrix[i][left]); } top++; bottom--; left++; right--; } return res; } };