valid-perfect-square

Author: bestgopher

README.md

@@ -7,14 +7,17 @@
 - 07：传递信息
   - [src](https://github.com/rustors/leetcode/blob/main/src/bin/chuan-di-xin-xi.rs)
   - [leetcode]((https://leetcode-cn.com/problems/chuan-di-xin-xi/))
-
 - 09：用两个栈实现队列
   - [src](https://github.com/rustors/leetcode/blob/main/src/bin/yong-liang-ge-zhan-shi-xian-dui-lie-lcof.rs)
   - [leetcode](https://leetcode-cn.com/problems/yong-liang-ge-zhan-shi-xian-dui-lie-lcof/)
 - 229：求众数 II
   - [src](https://github.com/rustors/leetcode/blob/main/src/bin/majority-element-ii.rs)
   - [leetcode](https://leetcode-cn.com/problems/majority-element-ii/)
+- 367：有效的完全平方数
+  - [src](https://github.com/rustors/leetcode/blob/main/src/bin/valid-perfect-square.rs)
+  - [leetcode](https://leetcode-cn.com/problems/valid-perfect-square/)
 - 387：字符串中的第一个唯一字符
+  
   - [src](https://github.com/rustors/leetcode/blob/main/src/bin/first-unique-character-in-a-string.rs)
   - [leetcode](https://leetcode-cn.com/problems/first-unique-character-in-a-string/)
 -  746：检查整数及其两倍数是否存在

src/bin/valid-perfect-square.rs

@@ -0,0 +1,48 @@
+fn main() {
+    // assert_eq!(true, Solution::is_perfect_square(4), "4");
+    // assert_eq!(false, Solution::is_perfect_square(5), "5");
+    // assert_eq!(false, Solution::is_perfect_square(14), "14");
+    // assert_eq!(true, Solution::is_perfect_square(16), "16");
+    // assert_eq!(false, Solution::is_perfect_square(2147483647), "2147483647");
+    assert_eq!(true, Solution::is_perfect_square(100000000), "100000000");
+}
+
+
+struct Solution;
+
+impl Solution {
+    // 使用二分的思维，
+    // 从1开始尝试，
+    // 当基数的平方小于num时，基数再次扩大2倍
+    // 当基数的平凡大于num时，则基数处于当前基数与上一个基数之间，取中间数尝试
+    // 当上一个基数与当前基数相等，且基数的平方不为num时，说明此数不是完全平方数，否则是完全平方数
+    pub fn is_perfect_square(num: i32) -> bool {
+        if num == 1 && num == 0 {
+            return true;
+        }
+
+        let mut num1 = 1;
+        let mut last_num = 1;
+
+        loop {
+            if num / num1 == num1 {
+                return if num % num1 == 0 {
+                    true
+                } else {
+                    false
+                };
+            } else if num / num1 > num1 {
+                last_num = num1;
+                num1 *= 2;
+            } else {
+                num1 = (last_num + num1) / 2;
+            }
+
+            if num1 == last_num && num / num1 != num1 {
+                return false;
+            }
+        }
+
+        false
+    }
+}