[#498] docs(FindDiagonalOrder) : add solution note

202508/498 FindDiagonalOrder/solution.md

@@ -0,0 +1,7 @@
+1. 觀察矩陣對角線模式
+   1. (0,0) -> 1 -> i+j = 0
+   2. (1, 0), (0, 1) -> 2 -> i+j = 1
+   3. (2, 0), (1, 1), (0, 2) -> 3 -> i+j = 2
+2. how to deal with diagonal sorting?
+   1. while (i+j) % 2 == 1 
+   2. while (i+j) % 2 == 0 reversed