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Author: assassint2017

Sum of Square Numbers/Sum_of_Square_Numbers.cpp

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+// 最暴力的方式就是n2的时间复杂度，但是可以用双指针加速到n的复杂度
+// 非常尴尬的是这样速度依然还是超时了，而且leftPtr如果使用int的话，也很容易发生溢出，因此需要使用long long
+class Solution 
+{
+public:
+    bool judgeSquareSum(int c) 
+    {
+        // 边界条件处理
+        if (c == 0 || c == 1)
+            return true;
+        
+        long long leftPtr = 0;
+        long long rightPtr = c;
+        long long res = c;
+        
+        while (leftPtr <= rightPtr)
+        {
+            if ((leftPtr * leftPtr + rightPtr * rightPtr) > res)
+                --rightPtr;
+            else if ((leftPtr * leftPtr + rightPtr * rightPtr) < res)
+                ++leftPtr;
+            else
+                return true;
+        }
+        
+        return false;
+    }
+};
+
+
+// 其实上边的思路是正确是，只要加上一个简单的trick就可以达到logn的时间复杂度，就是限制rightPtr的最大值
+// 所以这道题目最棒的一点就是需要考虑这个加速的小trick、以及使用更大的数据范围！！！！考察点还是非常棒的
+// Runtime: 0 ms, faster than 100.00% of C++ online submissions for Sum of Square Numbers.
+// Memory Usage: 8 MB, less than 100.00% of C++ online submissions for Sum of Square Numbers.
+
+class Solution 
+{
+public:
+    bool judgeSquareSum(int c) 
+    {
+        // 边界条件处理
+        if (c == 0 || c == 1)
+            return true;
+        
+        long long leftPtr = 0;
+        long long rightPtr = static_cast<int>(sqrt(c));
+        long long res = c;
+        
+        while (leftPtr <= rightPtr)
+        {
+            if ((leftPtr * leftPtr + rightPtr * rightPtr) > res)
+                --rightPtr;
+            else if ((leftPtr * leftPtr + rightPtr * rightPtr) < res)
+                ++leftPtr;
+            else
+                return true;
+        }
+        
+        return false;
+    }
+};

Sum of Square Numbers/Sum_of_Square_Numbers.py

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+# Runtime: 416 ms, faster than 15.92% of Python3 online submissions for Sum of Square Numbers.
+# Memory Usage: 13.8 MB, less than 50.00% of Python3 online submissions for Sum of Square Numbers.
+
+import math
+
+class Solution:
+    def judgeSquareSum(self, c: int) -> bool:
+        
+        if c is 0 or c is 1:
+            return True
+        
+        leftPtr = 0
+        rightPtr = int(math.sqrt(c))
+
+        while leftPtr <= rightPtr:
+            if leftPtr ** 2 + rightPtr ** 2 > c:
+                rightPtr -= 1
+            elif leftPtr ** 2 + rightPtr ** 2 < c:
+                leftPtr += 1
+            else:
+                return True
+        
+        return False