[LeetCode] 1090. Largest Values From Labels

使用 Java 爬取 LeetCode 题目内容以及提交的AC代码

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Description

We have a set of items: the i-th item has value values[i] and label labels[i].

Then, we choose a subset S of these items, such that:

• |S| <= num_wanted
• For every label L, the number of items in S with label L is <= use_limit.

Return the largest possible sum of the subset S.

 

Example 1:
Input: values = [5,4,3,2,1], labels = [1,1,2,2,3], num_wanted = 3, use_limit = 1
Output: 9
Explanation: The subset chosen is the first, third, and fifth item.

Example 2:
Input: values = [5,4,3,2,1], labels = [1,3,3,3,2], num_wanted = 3, use_limit = 2
Output: 12
Explanation: The subset chosen is the first, second, and third item.

Example 3:
Input: values = [9,8,8,7,6], labels = [0,0,0,1,1], num_wanted = 3, use_limit = 1
Output: 16
Explanation: The subset chosen is the first and fourth item.

Example 4:
Input: values = [9,8,8,7,6], labels = [0,0,0,1,1], num_wanted = 3, use_limit = 2
Output: 24
Explanation: The subset chosen is the first, second, and fourth item.

 

Note:

1. 1 <= values.length == labels.length <= 20000
2. 0 <= values[i], labels[i] <= 20000
3. 1 <= num_wanted, use_limit <= values.length

 

思路

题意：选取一个子集，子集中元素的总个数不大于 num_wanted，对于 label 值相同的元素，选取的个数不能大于 use_limit（如use_limit = 2，label值为 1 的元素有4个，最多选两个），这样的子集要求其 value 总和最大

题解：贪心，按照元素的value值排序，数值大的先选

 

+ View Code?

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static const auto io_sync_off = []() {     // turn off sync     std::ios::sync_with_stdio(false);     // untie in/out streams     std::cin.tie(nullptr);     return nullptr; }();   struct Node {     int value, label;       bool operator < (const Node &node)const{         return value > node.value;     } };   class Solution { public:       int largestValsFromLabels(vector<int> &values, vector<int> &labels, int num_wanted, int use_limit) {         int size = values.size();         Node node[size + 5];         int sum = 0;         map<int, int>mp;         map<int, int>::iterator it;         for (int i = 0; i < size; i++) {             node[i].value = values[i];             node[i].label = labels[i];         }         sort(node, node + size);         for (int i = 0; i < size && num_wanted; i++) {             if (mp.find(node[i].label) != mp.end()) {                 if (mp[node[i].label] < use_limit) {                     sum += node[i].value;                     mp[node[i].label]++;                     num_wanted--;                 }             } else {                 mp[node[i].label] = 1;                 sum += node[i].value;                 num_wanted--;             }         }         return sum;     } };