docs: add a description of the solution to lcci problem: No.08.05 (#761)

No.08.05.Recursive Mulitply

Author: YangFong

lcci/08.05.Recursive Mulitply/README.md

@@ -31,6 +31,59 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+~~**最佳方案：**~~
+
+~~直接返回 `A * B`~~
+
+正常递归，叠加总和
+
+```txt
+MULTIPLY(A, B)
+    if A == 0 || B == 0
+        return 0
+    A + multiply(A, B - 1)
+```
+
+优化 1：
+
+由数值较小的数字决定递归层次
+
+```txt
+MULTIPLY(A, B)
+    if A == 0 || B == 0
+        return 0
+    return max(A, B) + multiply(max(A, B), min(A, B) - 1)
+```
+
+优化 2：
+
+使用位移减少递归层次
+
+```txt
+MULTIPLY(A, B)
+    return (B % 1 == 1 ? A : 0) + (B > 1 ? MULTIPLY(A + A, B >> 1) : 0)
+```
+
+可进一步，转换为循环，虽然并不符合递归主题。
+
+> A 与 B 皆为**正整数**，初始值不会为 0，所以终止条件是 `B != 1`
+
+```txt
+MULTIPLY(A, B)
+    T = min(A, B)
+    A = max(A, B)
+    B = T
+    r = 0
+    while B != 1 {
+        if B % 2 == 1 {
+            r = r + A
+        }
+        A = A + A
+        B = B >> 1
+    }
+    return res + A
+```
+
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 ### **Python3**