Leetcode Maximum Product Subarray

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Leetcode Maximun Product Subarray 相关算法实现，使用dp算法完成问题，并提供测试，提供两种实现，一种为常规的，别一种为最大子串积的快速解决方法，代码如下。

常规解法：O(n^2)

#include <iostream>#include <vector>using namespace std;class Solution {public:    int maxProduct(vector<int>& nums) {        int len = nums.size();        vector<vector<vector<int> > > re(len, vector<vector<int> >(len, vector<int>(2, 0)));        re.resize(len);        for (int i = 0; i < len; i ++) {            re[i][i][0] = nums[i];            re[i][i][1] = nums[i];        }        for (int i = 1; i < len; i ++) {            for (int j = 0; j < len - i; j ++ ) {                int max = re[j][j][0];                for (int k = 0; k < i; k ++) {                    int temp = re[j][j + k][1] * re[j + k + 1][j + i][1];                    max = maxOfFour(max, temp, re[j][j + k][0], \                            re[j + k + 1][j + i][0]);                }                re[j][j + i][0] = max;                re[j][j + i][1] = re[j][j][1] * re[j + 1][j + i][1];            }        }        return re[0][len - 1][0];    }    int maxOfFour(int a, int b, int c, int d) {        int max = a;        if (b > max) {            max = b;        }        if (c > max) {            max = c;        }        if (d > max) {            max = d;        }        return max;    }};int main() {    vector<int> a;    a.push_back(-1);    a.push_back(2);    a.push_back(0);    a.push_back(-4);    a.push_back(3);    a.push_back(-2);    Solution so;    int re = so.maxProduct(a);    cout<<"The result is :"<<re<<endl;    return 0;}

由本问题特性实现的算法：O(n)

#include<iostream>#include<vector>#include<algorithm>using namespace std;class Solution {public:    int maxProduct(vector<int>& nums) {        int len = nums.size();        int reMax = nums[0];        int reMin = nums[0];        int re = nums[0];        for (int i = 1; i < len; i ++) {            int temp = reMax;            reMax = max(max(nums[i], reMin * nums[i]), reMax * nums[i]);            reMin = min(min(nums[i], reMin * nums[i]), temp * nums[i]);            re = max(reMax, re);        }        return re;    }};int main() {    vector<int> a;    a.push_back(-1);    a.push_back(0);    a.push_back(3);    a.push_back(-4);    Solution so;    int re = so.maxProduct(a);    cout<<"The result is :"<<re<<endl;    return 0;}

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