LeetCode spiral-marix

给定一个包含 m x n 个元素的矩阵(m 行, n 列),请按照顺时针螺旋顺序,返回矩阵中的所有元素。 示例 1:

输入: [ [ 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: [ [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]

解答:

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class Solution {
public:
vector<int> spiralOrder(vector<vector<int>>& matrix) {
if(matrix.empty() || matrix[0].empty()) return {};
vector<int> res;
int m = matrix.size(), n = matrix[0].size();
// 确定上下左右四条边的位置
int up = 0, down = m - 1, left = 0, right = n - 1;
while (true) {
for (int i = left; i <= right; i++) res.push_back(matrix[up][i]);
if (++up > down) break;
for (int i = up; i <= down; i++) res.push_back(matrix[i][right]);
if (--right < left) break;
for (int i = right; i >= left; i--) res.push_back(matrix[down][i]);
if (--down < up) break;
for (int i = down; i >= up; i--) res.push_back(matrix[i][left]);
if (++left > right) break;
}
return res;
}
};

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