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Author: assassint2017

Maximum Product of Three Numbers/Maximum_Product_of_Three_Numbers.cpp

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+// 第一种方式 线性的时间复杂度 手动维度若干个变量来记录最大值和最小值
+// Runtime: 48 ms, faster than 81.12% of C++ online submissions for Maximum Product of Three Numbers.
+// Memory Usage: 10.9 MB, less than 86.67% of C++ online submissions for Maximum Product of Three Numbers.
+
+class Solution 
+{
+public:
+    int maximumProduct(vector<int>& nums) 
+    {
+        int max1 = INT_MIN;
+        int max2 = INT_MIN;
+        int max3 = INT_MIN;
+
+        int min1 = INT_MAX;
+        int min2 = INT_MAX;
+
+        for (auto num : nums)
+        {
+            // 手动更新三个最大值
+            if (num >= max1)
+            {
+                max3 = max2;
+                max2 = max1;
+                max1 = num;
+            }
+            else if (num >= max2)
+            {
+                max3 = max2;
+                max2 = num;
+            }
+            else if (num > max3)
+                max3 = num;
+
+            // 手动更新两个最小值
+            if (num <= min1)
+            {
+                min2 = min1;
+                min1 = num;
+            }
+            else if (num < min2)
+                min2 = num;
+        }
+
+        int res1 = max1 * max2 * max3;
+        int res2 = max1 * min1 * min2;
+        return max(res1, res2);
+    }
+};
+
+
+// 如果题目问最大的N个数相乘的话，手动维护变量就不太方便了，因此可以考虑使用最大堆和最小堆的方式
+// Runtime: 48 ms, faster than 81.12% of C++ online submissions for Maximum Product of Three Numbers.
+// Memory Usage: 11 MB, less than 66.67% of C++ online submissions for Maximum Product of Three Numbers.
+class Solution 
+{
+public:
+    int maximumProduct(vector<int>& nums) 
+    {
+        // 初始化最小堆和最大堆
+        vector<int> maxHeap;
+        vector<int> minHeap;
+
+        for (int i = 0; i < 3; ++i)
+        {
+            maxHeap.push_back(nums[i]);
+            minHeap.push_back(nums[i]);
+        }
+        
+        sort(maxHeap.begin(), maxHeap.end());
+        maxHeap.pop_back();
+
+        auto great = [](const int& item1, const int& item2){return item1 > item2;};
+        make_heap(minHeap.begin(), minHeap.end(), great);
+
+        // 循环插入
+        for (int i = 3; i < nums.size(); ++i)
+        {
+            if (nums[i] < maxHeap[0])
+            {
+                pop_heap(maxHeap.begin(), maxHeap.end());
+                maxHeap.pop_back();
+
+                maxHeap.push_back(nums[i]);
+                push_heap(maxHeap.begin(), maxHeap.end());
+            }
+
+            if (nums[i] > minHeap[0])
+            {
+                pop_heap(minHeap.begin(), minHeap.end(), great);
+                minHeap.pop_back();
+
+                minHeap.push_back(nums[i]);
+                push_heap(minHeap.begin(), minHeap.end(), great);
+            }
+        }
+
+        // 计算最终结果
+        int res1 = minHeap[0] * minHeap[1] * minHeap[2];
+        int res2 = max(minHeap[1], minHeap[2]) * maxHeap[0] * maxHeap[1];
+        return max(res1, res2);
+    }
+};