Delete and Earn

题目来源：

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题目描述：

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].

心路历程：

这道题是leetcode上面位于动态规划目录下的一道题，难度为Medium。刚开始看到这道题的时候，没什么思路，找不到很好的子问题。所以写了一个遍历并递归的算法，输入测试用例倒没什么问题，但是提交后运行只通过了10个测试样例，后面的全部超时。如下图：

于是看了答案，确实有一种醍醐灌顶的感觉。先把所有的数从小到大排个序（因为数的范围有限，所以可以使用桶排序）。这样，问题就变成从左往右依次考虑是否加入某个数。过程中使用use和avoid两个变量来分别记录加入和不加入当前数所得到的值。接着考虑下一个数，若与前一个数不连续。可以直接加入；若与前一个数连续，就要加上前一个数的avoid值。这里有一点要注意，每个数都有可能重复，但是它们要不全部加入，要么全部不加，这是因为当加入第一个这种数时，它的左右邻都会被删除，所以其它的相同的数没办法再被删除（因为没有邻居了）。leetcode上给了Java和Python的解法，下面给出了类似的C++解法。

解决方案：

#include<iostream>#include<vector>#include<algorithm>using namespace std;class Solution {public:    int deleteAndEarn(vector<int>& nums) {        int count[10001] = { 0 };        for (int i = 0; i < nums.size(); i++) count[nums[i]]++;        int avoid = 0;        int use = 0;        int prev = -1;        for (int k = 0; k <= 10000; ++k) {            if (count[k] > 0) {                int m = max(avoid, use);                if (k - 1 != prev) {                    use = k * count[k] + m;                    avoid = m;                }                else {                    use = k * count[k] + avoid;                    avoid = m;                }                prev = k;            }        }        return max(avoid, use);    }};int main() {    vector<int> nums;    nums.push_back(3);    nums.push_back(4);    nums.push_back(2);    Solution solution;    int result = solution.deleteAndEarn(nums);    cout << result<<endl;    system("pause");    return 0;}