LeetCode_Jackis

problems/0152_Maximum_Product_Subarray/Jackis.cpp

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+//1
+class Solution {
+public:
+    int maxProduct(vector<int>& nums) {
+        //dp Solution
+        int n = nums.size();
+        vector<int> f(n);      //max
+        vector<int> g(n);      //min
+        int res;
+        res = f[0] = g[0] = nums[0];
+        for(int i=1; i<n; i++)
+        {
+            f[i] = max(max(f[i-1]*nums[i], g[i-1]*nums[i]), nums[i]);
+            g[i] = min(min(f[i-1]*nums[i], g[i-1]*nums[i]), nums[i]);
+            if(f[i] > res)
+                res = f[i];
+        }
+        return res;
+    }
+};
+
+//2
+class Solution {
+public:
+    int maxProduct(vector<int>& nums) {
+        //dp Solution
+       int res=nums[0], mx=res, mn=res;
+        for(int i=1; i<nums.size(); i++)
+        {
+            if(nums[i] > 0)
+            {
+                mx = max(mx*nums[i], nums[i]);
+                mn = min(mn*nums[i], nums[i]);
+            }
+            else 
+            {
+                int tmp = mx;
+                mx = max(mn*nums[i], nums[i]);
+                mn = min(tmp*nums[i], nums[i]);
+            }
+            res = max(res, mx);
+        }
+        return res;
+    }
+};
+//3
+class Solution {
+public:
+    int maxProduct(vector<int>& nums) {
+        //dp Solution
+       int res=nums[0], mx=res, mn=res;
+        for(int i=1; i<nums.size(); i++)
+        {
+            if(nums[i] < 0)
+                swap(mx, mn);
+            mx = max(mx*nums[i], nums[i]);
+            mn = min(mn*nums[i], nums[i]);
+            res = max(res, mx);
+        }
+        return res;
+    }
+};
+//4
+class Solution {
+public:
+    int maxProduct(vector<int>& nums) {
+       int res=nums[0], prod=1, n=nums.size();
+        for(int i=0; i<n; i++)
+        {
+            res = max(res, prod*=nums[i]);
+            if(nums[i] == 0)
+                prod = 1;
+        }
+        prod = 1;
+        for(int i=n-1; i>=0; i--)
+        {
+            res = max(res, prod*=nums[i]);
+            if(nums[i] == 0)
+                prod = 1;
+        }
+        return res;
+    }
+};

problems/0973_K_Closest_Points_to_Origin/Jackis.c

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+/*
+基于选择排序的方法，但该方法提交时显示超时
+*/
+
+class Solution {
+public:
+    vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
+        vector<vector<int>> res;
+        SelectSort(points, res, K);
+        return res;
+    }
+    double Distance(vector<int> n)
+    {
+        return (double)n[0]*n[0] + (double)n[1]*n[1];
+    }  
+    void SelectSort(vector<vector<int>>& num, vector<vector<int>>& res, int K)
+    {
+        double min;
+        int min_pos;
+        int i, j;
+        for(i=0; i<K; i++)
+        {
+            min = Distance(num[i]);
+            min_pos = i;
+            for(j=i+1; j<num.size(); ++j)
+            {
+                double num_j = Distance(num[j]);
+                if(num_j < min)
+                {
+                    min = num_j;
+                    min_pos = j;
+                }
+            }
+            swap(num[i], num[min_pos]);
+            res.push_back(num[i]);
+        }
+
+    }
+};
+/*
+直接使用STL里的sort函数排序，只保留下标为[0--k-1]的数
+*/
+class Solution {
+public:
+    vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
+       sort(begin(points), end(points), [this](auto& f, auto& s)
+        {
+            return (f[0]*f[0] + f[1]*f[1] < s[0]*s[0] + s[1]*s[1]); 
+        });
+
+       points.erase(begin(points) + K, end(points));
+       return points;
+    }
+};

submittors/Jackis-1.txt

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+This is