动态规划问题的解决步骤

leetcode198 House Robber

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.

class Solution { /*       1.暴力搜索 O(2^n)    int solve(int idx, vector<int>& nums)    {           if(idx < 0)            return 0;        if(idx == 0)            return nums[0];        else if(idx == 1)            return max(nums[0], nums[1]);        return max( solve(idx-2, nums) + nums[idx], solve(idx - 1, nums));    }     int rob(vector<int>& nums)     {        int n = nums.size();        return solve(n-1, nums);    }    */    /*    2. 以空间换时间， O(n)private:    vector<int> result;private:    int solve(int idx, vector<int>& nums)    {        if(idx < 0)            return 0;        if(result[idx] >= 0)            return result[idx];        result[idx] = max(solve(idx - 2, nums) + nums[idx], solve(idx - 1, nums));        return result[idx];    }public:    int rob(vector<int>& nums)     {        int n = nums.size();        if(n==0)            return 0;        result.resize(n);        for(int i=0; i < n; ++i)        {            result[i] = -1;        }        return solve(n-1, nums);    }    *//*3. 递归转迭代，变身动态规划private:    vector<int> result;public:    int rob(vector<int>& nums) {        int n = nums.size();        if(n==0)            return 0;        result.resize(n);        if(n >= 1)            result[0] = nums[0];        if(n >= 2)            result[1] = max(nums[0], nums[1]);        for(int i = 2; i < n; ++i)        {            result[i] = max(result[i - 2] + nums[i], result[i - 1]);        }        return result[n-1];    }*/    // 滚动数组 空间最小化 O(n) ==>> O(3)public:    int rob(vector<int>& nums) {        int n = nums.size();        if(n==0)            return 0;        vector<int> result(3);        if(n >= 1)            result[0] = nums[0];        if(n >= 2)            result[1] = max(nums[0], nums[1]);        for(int i = 2; i < n; ++i)        {            result[i % 3] = max(result[ (i - 2) % 3] + nums[i], result[ (i - 1) % 3]);        }        return result[ (n-1) % 3];    }};