LeetCode#740 Delete and Earn (week16)

week16

题目

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.

Note:

· The length of nums is at most 20000.
· Each element nums[i] is an integer in the range [1, 10000].
原题地址：https://leetcode.com/problems/delete-and-earn/description/

解析

题目给定一些数（范围在1到10000），如果删除某个数n，则可以earn到这个数但值为n-1和n+1的数会被删除，求能够earn到的最大的值。
思路：可以用一般的动态规划思想来解决这个问题，从值1开始，计算到目前为止能earn到的最大值，对于某个值n，对应的最大值可由公式f(n)=max{f(n-1)，f(n-2)+n*numsOf(n)}得到，前者表示取n-1（则n会被删除），后者表示取n（则n-1会被删除），从1开始循环执行最终到10000得到最后结果。

代码

class Solution {public:    int deleteAndEarn(vector<int>& nums) {        vector<int> count(10001, 0);        for (int i = 0; i < nums.size(); ++i) {            ++count[nums[i]];        }        int *earn = new int[10001];        earn[1] = count[1];        for (int i = 2; i < 10001; ++i) {            earn[i] = max(earn[i - 1], earn[i - 2] + count[i] * i);        }        return earn[10000];    }    int max(int a, int b) {        return a > b ? a : b;    }};