diff --git a/矩阵/1. 螺旋矩阵.md b/矩阵/1. 螺旋矩阵.md new file mode 100644 index 0000000..f80e9ed --- /dev/null +++ b/矩阵/1. 螺旋矩阵.md @@ -0,0 +1,83 @@ +# 螺旋矩阵 + +## 题目 + +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。 + + + +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 + +## 思路 + +- 模拟 + - 方向 + - →↓←↑ + - 边界 + - `up` + - `down` + - `left` + - `right` + +## 代码 + +```java +class Solution { + public List spiralOrder(int[][] matrix) { + // Init: up, down, left, right + int up = 0; + int down = matrix.length - 1; + int left = 0; + int right = matrix[0].length - 1; + // Init: ans + List ans = new ArrayList<>(); + int tar = matrix.length * matrix[0].length; + // Traverse + while (left <= right && up <= down) { + for (int i = left; i <= right; i++) { + ans.add(matrix[up][i]); + } + up ++; + + for (int i = up; i <= down; i++) { + ans.add(matrix[i][right]); + } + right --; + + if (up <= down) { + for (int i = right; i >= left; i--) { + ans.add(matrix[down][i]); + } + } + down --; + + if (left <= right) { + for (int i = down; i >= up; i--) { + ans.add(matrix[i][left]); + } + } + left ++; + + } + + return ans; + } +} +``` \ No newline at end of file