Site updated: 2022-08-28 22:30:16

baidusitemap.xml

@@ -89,11 +89,11 @@
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+    <loc>https://nicksxs.me/2021/10/03/%E8%81%8A%E4%B8%80%E4%B8%8B-RocketMQ-%E7%9A%84%E6%B6%88%E6%81%AF%E5%AD%98%E5%82%A8%E4%B8%89/</loc>
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@@ -656,6 +656,10 @@
     <loc>https://nicksxs.me/2015/01/04/Path-Sum/</loc>
     <lastmod>2020-01-12</lastmod>
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+  <url>
+    <loc>https://nicksxs.me/2015/06/22/invert-binary-tree/</loc>
+    <lastmod>2020-01-12</lastmod>
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     <lastmod>2019-12-22</lastmod>

code/Solution1260.java

@@ -0,0 +1,44 @@
+public class Solution1260 {
+
+  public List<List<Integer>> shiftGrid(int[][] grid, int k) {
+    // 行数
+    int m = grid.length;
+    // 列数
+    int n = grid[0].length;
+    // 偏移值，取下模
+    k = k % (m * n);
+    // 反向取下数量，因为我打算直接从头填充新的矩阵
+    /*
+     *    比如
+     *    1 2 3
+     *    4 5 6
+     *    7 8 9
+     *    需要变成
+     *    9 1 2
+     *    3 4 5
+     *    6 7 8
+     *    就要从 9 开始填充
+     */
+    int reverseK = m * n - k;
+    List<List<Integer>> matrix = new ArrayList<>();
+    // 这类就是两层循环
+    for (int i = 0; i < m; i++) {
+      List<Integer> line = new ArrayList<>();
+      for (int j = 0; j < n; j++) {
+        // 数量会随着循环迭代增长, 确认是第几个
+        int currentNum = reverseK + i * n + (j + 1);
+        // 这里处理下到达矩阵末尾后减掉 m * n
+        if (currentNum > m * n) {
+          currentNum -= m * n;
+        }
+        // 根据矩阵列数 n 算出在原来矩阵的位置
+        int last = (currentNum - 1) % n;
+        int passLine = (currentNum - 1) / n;
+
+        line.add(grid[passLine][last]);
+      }
+      matrix.add(line);
+    }
+    return matrix;
+  }
+}