**Leetcode 740. Delete and Earn

https://leetcode.com/problems/delete-and-earn/description/

跟买卖股票的还是一样的思路。

参考率 https://leetcode.com/problems/delete-and-earn/discuss/123634/C++-9ms-DP-solution

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        int n = nums.size();
        if (n == 0) return 0;
        vector< vector<int> > dp( nums.size() + 1, vector<int>(2) );
        sort(nums.begin(), nums.end());
        // 0表示不要这个数，1表示要这个数
        
        dp[0][0] = 0;
        dp[0][1] = nums[0];
        for (int i = 1; i < nums.size(); i++) {
            if (nums[i] == nums[i-1]) {
                dp[i][0] = dp[i-1][0];
                dp[i][1] = dp[i-1][1] + nums[i];
                continue;
            }
            dp[i][0] = max(dp[i-1][0], dp[i-1][1]);
            dp[i][1] = nums[i] == nums[i-1] + 1 ? dp[i-1][0] + nums[i] : dp[i][0] + nums[i];
        }
        return max(dp[n-1][0], dp[n-1][1]);
    }
};

https://leetcode.com/problems/delete-and-earn/discuss/109871/Awesome-Python-4-liner-with-explanation-Reduce-to-House-Robbers-Question

这里有个非常好的 hash转化成house robber的做法。当时我想到这个题感觉跟不能取相邻的XXX的感觉一样，但是没想到咋做。。

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        vector<int> dp(10000+1, 0);
        for (int i  = 0; i < nums.size(); i++) {
            dp[ nums[i] ] += nums[i];
        }
        
        int last = 0, cur = 0;
        for (int i = 0; i < dp.size(); i++) {
            // cur其实是选了i-1的，last是cur之前的选择
            int tmp = cur;
            cur = max(last + dp[i], cur);
            last = tmp;
        }
        return cur;
    }
};

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