[leetcode] 740. Delete and Earn

Question:

Given an array nums of integers, you can perform operations on the array.

In each operation, you pick any nums[i] and delete it to earn nums[i] points. After, you must delete every element equal to nums[i] - 1 or nums[i] + 1.

You start with 0 points. Return the maximum number of points you can earn by applying such operations.

Example 1:

Input: nums = [3, 4, 2]Output: 6

Explanation:
Delete 4 to earn 4 points, consequently 3 is also deleted.
Then, delete 2 to earn 2 points. 6 total points are earned.

Example 2:

Input: nums = [2, 2, 3, 3, 3, 4]Output: 9

Explanation:
Delete 3 to earn 3 points, deleting both 2’s and the 4.
Then, delete 3 again to earn 3 points, and 3 again to earn 3 points.
9 total points are earned.
Note:

• The length of nums is at most 20000.
• Each element nums[i] is an integer in the range [1, 10000].

Solution:

看了discuss才会做。就是把原问题归约到以前做过的一道题198. House Robber

归约的过程如下：
1. 把原数组出现的数对应于新数组的下标
2. 原数组的一个数出现一次，则新数组对应下标的数自增该下标的值
3. 这样就构造出了一个新的数组，对应于House Robber的每个房子价值的数组
4. 这样，按照规定，不能抢相邻房子，即不能同时取数组相邻的数

为什么这样是对的呢，因为，当我们一旦决定要取一个数 i，则不能再取i+1或i-1了，当做了上述转换后，每个“房子”就对应着一个数 i，我们不能抢相邻房子就等同于不能取了 i 后还想着取 i+1 或 i-1。

然后应用House Robber的解法即可。

class Solution {public:    int deleteAndEarn(vector<int>& nums) {        int size = 0;        for (int i : nums)            if (size < i)                size = i;        vector<int> house(size + 1, 0);        for (int i : nums)            house[i] += i;        vector<int> rob(size + 1, 0);        rob[1] = house[1];        for (int i = 2; i < house.size(); i++)            rob[i] = max(rob[i-1] - house[i-1], rob[i-2]) + house[i];        return max(rob[size], rob[size - 1]);    }};