一.题目要求

给你一个 m 行 n 列的矩阵 matrix ，请按照顺时针螺旋顺序 ，返回矩阵中的所有元素。

二.题目难度

中等

三.输入样例

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]

示例 2：

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]

提示：
m == matrix.length
n == matrix[i].length
1 <= m, n <= 10
-100 <= matrix[i][j] <= 100

四.解题思路

找到四个方向坐标变化的规律遍历即可

五.代码实现

优化后

class Solution {
public:vector<int> spiralOrder(vector<vector<int>>& matrix) {vector<int> sup;int width = matrix.size();int length = matrix[0].size();int left = 0;int right = length - 1;int top = 0;int bottom = width - 1;while (left <= right && top <= bottom){for (int i = left; i <= right; i++){sup.push_back(matrix[top][i]);}top++;for (int i = top; i <= bottom; i++){sup.push_back(matrix[i][right]);}right--;if (top <= bottom){for (int i = right; i >= left; i--){sup.push_back(matrix[bottom][i]);}bottom--;}if (left <= right){for (int i = bottom; i >= top; i--){sup.push_back(matrix[i][left]);}left++;}}return sup;}
};

优化前

class Solution {
public:vector<int> spiralOrder(vector<vector<int>>& matrix) {int wid = matrix.size();int len = matrix[0].size();int crtlen = len;int crtwid = wid;vector<int> ans;bool lefttoright = true;bool toptobottom = true;int lenindex = 0, widindex = 0, i;while (crtlen && crtwid){if (lefttoright){for (i = 0; i < crtlen; i++){ans.push_back(matrix[lenindex][widindex++]);}lenindex++; widindex--;lefttoright = !lefttoright;crtwid--;}else{for (i = 0; i < crtlen; i++){ans.push_back(matrix[lenindex][widindex--]);}lenindex--; widindex++;lefttoright = !lefttoright;crtwid--;}if (toptobottom){for (i = 0; i < crtwid; i++){ans.push_back(matrix[lenindex++][widindex]);}widindex--; lenindex--;toptobottom = !toptobottom;crtlen--;}else{for (i = 0; i < crtwid; i++){ans.push_back(matrix[lenindex--][widindex]);}widindex++;lenindex++;toptobottom = !toptobottom;crtlen--;}}return ans;}
};

六.题目总结

注意溢出判断