[Leetcode学习-c++&java]Reduce Array Size to The Half
2021/7/6 17:39:05
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问题:
难度:media
说明:
给出一个数组,里面包含重复元素,将任意种同值元素移除,剩下的数组元素长度 <= 原来数组的一半,求最小移除同值元素数量。
题目连接:https://leetcode.com/problems/reduce-array-size-to-the-half/
输入范围:
1 <= arr.length <= 10^5
arr.length
is even.1 <= arr[i] <= 10^5
输入案例:
Example 1: Input: arr = [3,3,3,3,5,5,5,2,2,7] Output: 2 Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array). Possible sets of size 2 are {3,5},{3,2},{5,2}. Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has size greater than half of the size of the old array. Example 2: Input: arr = [7,7,7,7,7,7] Output: 1 Explanation: The only possible set you can choose is {7}. This will make the new array empty. Example 3: Input: arr = [1,9] Output: 1 Example 4: Input: arr = [1000,1000,3,7] Output: 1 Example 5: Input: arr = [1,2,3,4,5,6,7,8,9,10] Output: 5
我的代码:
其实这个主要是审题,比如 [7,7,7,7] 就代表 4 个 7,那么我移除 7 ,数组长度就变成了 0,我移除的最小类型数量是 1,那么我返回 1,。
这个只需要把所有元素用hash处理一次,求出各种数值的统计,然后排序一次,从最大数量元素开始移除,移除到一半或以上就可以了。
用 HashMap 或者 数组也行
Java:
class Solution { private static int[] cache = new int[100001]; public int minSetSize(int[] arr) { for(int i : arr) cache[i] ++; Arrays.sort(cache); int total = 0, count = 0, half = (arr.length & 1) == 0 ? arr.length >> 1 : (arr.length >> 1) + 1; for(int i = 100000; total < half && i > 0; i --) { total += cache[i]; count ++; } Arrays.fill(cache, 0); return count; } public int minSetSize2(int[] arr) { Map<Integer, Integer> cache = new HashMap<>(); for(int i : arr) cache.put(i, cache.getOrDefault(i, 0) + 1); TreeMap<Integer, Integer> tree = new TreeMap<Integer, Integer>(new Comparator<Integer>() { public int compare(Integer o1, Integer o2) { return o2 - o1; } }); for(int i : cache.values()) tree.put(i, tree.containsKey(i) ? tree.get(i) + 1 : 1); int total = 0, count = 0, half = (arr.length & 1) == 0 ? arr.length >> 1 : (arr.length >> 1) + 1; while(!tree.isEmpty() && total < half) { Map.Entry<Integer, Integer> entry = tree.pollFirstEntry(); int times = entry.getValue(); while(total < half && times -- > 0) { total += entry.getKey(); count ++; } } return count; } }
C++:
static int cache[100001] = {0}; class Solution { public: int minSetSize(vector<int>& arr) { for(int i : arr) cache[i] ++; sort(cache, cache + 100001); int total = 0, count = 0, half = (arr.size() & 1) == 0 ? arr.size() >> 1 : (arr.size() >> 1) + 1; for(int i = 100000; total < half && i > 0; i --) { total += cache[i]; count ++; } memset(cache, 0, sizeof(cache)); return count; } };
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