- 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
numsis at most20000. - Each element
nums[i]is an integer in the range[1, 10000].
- 解题思路
看题目，可以意识到是动态规划类型的题目，但不知道怎么写迭代式子，就是相不清楚状态。所以，一开始心虚的按照自己的贪婪算法实现了下，结果就是代码极多，但结果不能对～
无奈，开始查阅资料，找到了一篇比较靠谱的博客。借鉴他的思路，自己努力写了下动态规划的实现。思路的关键在于：
- 取出一个数，其收益为 数的频数 × 数的值。按照规则，取出一个，必然取出该值的所有数。
- 两个状态，取出当前数的最大收益（maxFetch），不取当前数的最大收益（maxNoFetch）。
- 初始状态：
- maxFetch = 0, maxNoFetch = 0;
- 当前状态与上一状态的关系
- 不取当前数，则为上一状态的最大值（max(prevMaxFetch, prevMaxNoFetch)）。
- 取出当前数，若数和上一状态的数关联（+/- 1），则为prevMaxNoFetch + 取出数的收益。否则，为max(prevMaxFetch, prevMaxNoFetch) + 取出数的收益。
- 示例代码
class Solution {
public:
int deleteAndEarn(vector<int>& nums) {
map<int, int> freqs;
int size = nums.size();
for(int i = ; i < size; i++)
{
int curr = nums[i];
// remember frequences for curr
if(freqs.find(curr) == freqs.end())
{
freqs[curr] = ;
}
else
{
freqs[curr] += ;
}
}
int maxFetch = , maxNoFetch = ;
int prevMaxFetch = , prevMaxNoFetch = ;
map<int, int>::iterator prevChoice;
map<int, int>::iterator currChoice;
// calculate maximum according to previous status
for(currChoice = freqs.begin(); currChoice != freqs.end(); ++currChoice)
{
// initiate
if(currChoice == freqs.begin())
{
// get this number
maxFetch = currChoice->first * currChoice->second;
// do not get this number
maxNoFetch = ;
}
// transferring
else
{
prevMaxFetch = maxFetch;
prevMaxNoFetch = maxNoFetch;
// do not get the number
maxNoFetch = max(prevMaxFetch, prevMaxNoFetch);
// get this number
if(currChoice->first == prevChoice -> first + || currChoice->first == prevChoice -> first - ) {
// related -> must not fetch previous node
maxFetch = prevMaxNoFetch + currChoice->first * currChoice->second;
}
else
{ // non related
maxFetch = maxNoFetch + currChoice->first * currChoice->second;
}
}
prevChoice = currChoice;
}
return max(maxFetch, maxNoFetch);
}
};