分析

class Solution {
public:int deleteAndEarn(vector<int>& nums) {const int N = 10001;// 预处理int arr[N] = {0};for (int& e : nums)arr[e] += e;// 在 arr 上进行 打家劫舍 问题vector<int> f(N), g(N);// 填表for(int i = 1; i < N; i++){f[i] = g[i - 1] + arr[i];g[i] = max(f[i - 1], g[i - 1]);}return max(f[N - 1], g[N - 1]);}
};

分析：

class Solution {
public:int minCost(vector<vector<int>>& costs) {const int N = costs.size() + 1;vector<int> dp0(N), dp1(N), dp2(N);for (int i = 1; i < N; i++) {dp0[i] = costs[i - 1][0] + min(dp1[i - 1], dp2[i - 1]);dp1[i] = costs[i - 1][1] + min(dp0[i - 1], dp2[i - 1]);dp2[i] = costs[i - 1][2] + min(dp1[i - 1], dp0[i - 1]);}return min(dp0[N - 1], min(dp1[N - 1], dp2[N - 1]));}
};

class Solution {
public:int minCost(vector<vector<int>>& costs) {const int N = costs.size() + 1;vector<vector<int>> dp(N, vector<int>(3, 0));for (int i = 1; i < N; i++) {dp[i][0] = costs[i - 1][0] + min(dp[i - 1][1], dp[i - 1][2]);dp[i][1] = costs[i - 1][1] + min(dp[i - 1][0], dp[i - 1][2]);dp[i][2] = costs[i - 1][2] + min(dp[i - 1][1], dp[i - 1][0]);}return min(dp[N - 1][0], min(dp[N - 1][1], dp[N - 1][2]));}
};