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Author: assassint2017

Max Points on a Line/Max_Points_on_a_Line.cpp

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+// 这道题目很显然可以使用立方的时间复杂度，常数的空间复杂度来解决
+// 但是，可以使用哈希表牺牲空间来换取时间
+
+// Runtime: 24 ms, faster than 41.41% of C++ online submissions for Max Points on a Line.
+// Memory Usage: 12.6 MB, less than 24.54% of C++ online submissions for Max Points on a Line.
+
+class Solution 
+{
+public:
+    int maxPoints(vector<vector<int>>& points) 
+    {
+        // 边界条件处理
+        if (points.size() <= 2) return points.size();
+        
+        int res = 2;
+        for (int i = 0; i < points.size(); ++i)
+        {
+            int tempres = 0, overlap = 0;
+            unordered_map<string, int> hashMap;
+            
+            for (int j = i; j < points.size(); ++j)
+            {
+                if (points[i][0] == points[j][0] && points[i][1] == points[j][1])
+                {
+                    ++overlap;
+                    continue;
+                }
+                
+                string slope = getSlope(points[i], points[j]);    
+                if (hashMap.find(slope) == hashMap.end())
+                {
+                    hashMap.insert(pair<string, int>(slope, 1));
+                    tempres = max(tempres, 1);
+                }
+                else
+                    tempres = max(tempres, ++hashMap.at(slope));
+            }
+            res = max(res, tempres + overlap);
+        }
+        return res;
+    }
+private:
+    string getSlope(const vector<int>& point1, const vector<int>& point2)
+    {
+        int a = point1[1] - point2[1];
+        int b = point1[0] - point2[0];
+        
+        // 如果两点形成了一条垂线
+        if (b == 0) return to_string(point1[0]) + "*";
+        
+        // 如果两点之间形成了一条水平线
+        if (a == 0) return "*" + to_string(point1[1]);
+        
+        // 两点形成一条正常的线段
+        int gcd = getGcd(a, b);
+        a /= gcd, b /= gcd;
+        
+        if ((a > 0 && b < 0 )|| (a < 0 && b < 0))
+            a *= -1, b*= -1;
+        return to_string(a) + '*' + to_string(b);
+    }
+    
+    // 求取两数的最大公约数
+    int getGcd(int num1, int num2)
+    {
+        while (num1 % num2 != 0)
+        {
+            int temp = num1 % num2;
+            num1 = num2;
+            num2 = temp;
+        }
+        return num2;
+    }
+};