LC 740. Delete and Earn
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].
自以为做了一个区间DP,结果并不是这样做。
//
// Created by yuxi on 2019/1/22.
// #include <vector>
#include <unordered_map>
#include <algorithm>
using namespace std; class Solution {
public:
unordered_map<int,int> mp;
unordered_map<int,int> memo;
int deleteAndEarn(vector<int>& nums) {
if(nums.empty()) return ;
for(int x : nums) mp[x]++;
vector<int> keys;
for(auto it = mp.begin(); it != mp.end(); it++) keys.push_back(it->first);
sort(keys.begin(),keys.end());
vector<vector<int>> dp(keys.size(), vector<int>(keys.size(),));
for(int k=; k<keys.size(); k++) {
for(int i=; i<keys.size(); i++) {
int j = i - k;
if(j < ) continue;
if(i == j) {
dp[j][i] = keys[i] * mp[keys[i]];
continue;
}
for(int l = j; l < i; l++) {
if(keys[l] == keys[l+]-) {
if(l+ == i) dp[j][i] = max(dp[j][i], max(dp[j][l],dp[i][i]));
else dp[j][i] = max(dp[j][i], dp[j][l]+dp[l+][i]);
}else {
dp[j][i] = max(dp[j][i], dp[j][l]+dp[l+][i]);
}
}
}
}
return dp[][keys.size()-];
}
};
Runtime: 8 ms, faster than 87.68% of C++ online submissions for Delete and Earn.
//
// Created by yuxi on 2019/1/22.
// #include <vector>
#include <unordered_map>
#include <algorithm>
using namespace std; class Solution {
public:
unordered_map<int,int> mp;
unordered_map<int,int> memo;
int deleteAndEarn(vector<int>& nums) {
if(nums.empty()) return ;
int maxval = ;
for(int x : nums) {
maxval = max(maxval,x);
mp[x]++;
}
vector<int> a(maxval+, );
for(auto it=mp.begin(); it != mp.end(); it++) {
a[it->first] = it->second;
}
int prev = , cur = ;
for(int i=; i<a.size(); i++) {
cur = max(cur, i > ? a[i] * i + a[i-] : a[i] * i);
a[i] = cur;
}
return cur;
}
};
LC 740. Delete and Earn的更多相关文章
- 740. Delete and Earn
Given an array nums of integers, you can perform operations on the array. In each operation, you pic ...
- leetcode笔记(六)740. Delete and Earn
题目描述 Given an array nums of integers, you can perform operations on the array. In each operation, yo ...
- LeetCode 740. Delete and Earn
原题链接在这里:https://leetcode.com/problems/delete-and-earn/ 题目: Given an array nums of integers, you can ...
- 【leetcode】740. Delete and Earn
题目如下: Given an array nums of integers, you can perform operations on the array. In each operation, y ...
- [LeetCode]Delete and Earn题解(动态规划)
Delete and Earn Given an array nums of integers, you can perform operations on the array. In each op ...
- [LeetCode] Delete and Earn 删除与赚取
Given an array nums of integers, you can perform operations on the array. In each operation, you pic ...
- [Swift]LeetCode740. 删除与获得点数 | Delete and Earn
Given an array nums of integers, you can perform operations on the array. In each operation, you pic ...
- LC 450. Delete Node in a BST
Given a root node reference of a BST and a key, delete the node with the given key in the BST. Retur ...
- LC 955. Delete Columns to Make Sorted II
We are given an array A of N lowercase letter strings, all of the same length. Now, we may choose an ...
随机推荐
- 【Day3】1.正则表达式
1.正则表达式 2.案例 关闭贪婪模式
- 8. Object References, Mutability, and Recycling
1. Variables Are Not Boxes # Think variables as sticky notes a = [1, 2, 3] b = a a.append(4) print b ...
- 在DjangoAdmin中使用KindEditor(上传图片)
一.下载 http://kindeditor.net/down.php 删除asp.asp.net.php.jsp.examples文件夹 拷贝到static目录下 二.配置 kindeditor目录 ...
- ReaderWriterLockSlim使用示例
/// <summary> /// ReaderWriterLockSlim使用示例 /// </summary> internal sealed class Transact ...
- C/C++代码规范
零.前言 笔者最近在看开源代码,看到代码格式各自参差不齐,感觉像是各家各有所长.因此打算写一篇关于C/C++代码规范文章,请各位参考,并践踏批评. 一.文件排版 1. 包含头文件 • 先系统头文件,后 ...
- Codeforces 1175F The Number of Subpermutations
做法①:RMQ(预处理NLOGN+后续跳跃蜜汁复杂度) 满足题意的区间的条件转换: 1.长度为R-L+1则最大值也为R-L+1 2.区间内的数不重复 当RMQ(L,R)!=R-L+1时 因为已经保证了 ...
- C++-cin与scanf cout与printf效率问题
http://blog.csdn.net/l2580258/article/details/51319387 void cin_read_nosync() { freopen("data.t ...
- js动画fireworks烟花
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...
- 域知识深入学习一:Active Directory 域服务
AD DS用来组织,管理,控制网络资源 1.1 Active Directory 域服务概述 AD内的directorydatabase(目录数据库)用来存储用户账户,计算机账户,打印机与共享文件 ...
- [六省联考2017]分手是祝愿——期望DP
原题戳这里 首先可以确定的是最优策略一定是从大到小开始,遇到亮的就关掉,因此我们可以\(O(nlogn)\)的预处理出初始局面需要的最小操作次数\(tot\). 然后容(hen)易(nan)发现即使加 ...