数组去重和排序 (100%用例)C卷 (Java&&Python&&Node.js&&C语言&&C++)
2023-12-28 13:12:35
给定一个乱序的数组,删除所有的重复元素,使得每个元素只出现一次,并且按照出现的次数从高到低进行排序,相同出现次数按照第一次出现顺序进行先后排序。
输入描述:
一个数组
输出描述:
去重排序后的数组
示例1?输入输出示例仅供调试,后台判题数据一般不包含示例
输入
1,3,3,3,2,4,4,4,5
输出
3,4,1,2,5
备注:
数组大小不超过100
数组元素值大小不超过100
Java版本
import java.util.*;
public class Main{
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
String input = scanner.nextLine();
String[] numbers = input.split(",");
List<Integer> s = new ArrayList<>();
for (String number : numbers) {
s.add(Integer.parseInt(number));
}
Map<Integer, Integer> freqMap = new HashMap<>();
for (int number : s) {
freqMap.put(number, freqMap.getOrDefault(number, 0) + 1);
}
List<Map.Entry<Integer, Integer>> freqList = new ArrayList<>(freqMap.entrySet());
Collections.sort(freqList, new Comparator<Map.Entry<Integer, Integer>>() {
@Override
public int compare(Map.Entry<Integer, Integer> o1, Map.Entry<Integer, Integer> o2) {
return o2.getValue().compareTo(o1.getValue());
}
});
StringBuilder result = new StringBuilder();
for (Map.Entry<Integer, Integer> entry : freqList) {
result.append(entry.getKey()).append(",");
}
System.out.println(result.toString().substring(0, result.length() - 1));
}
}
Python版本
from collections import Counter
if __name__ == '__main__':
s = [int(i) for i in input().split(',')]
dic = dict(Counter(s))
dic = dict(sorted(dic.items(), key=lambda x: -x[1]))
print(",".join([str(i) for i in dic.keys()]))
C++版本
#include <iostream>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
string input;
getline(cin, input);
vector<int> s;
int value;
size_t start_pos = 0;
while ((start_pos = input.find(',')) != string::npos) {
value = stoi(input.substr(0, start_pos));
s.push_back(value);
input.erase(0, start_pos + 1);
}
s.push_back(stoi(input)); // 加入最后一个数值
map<int, int> freqMap;
for (int i = 0; i < s.size(); i++) {
freqMap[s[i]]++;
}
vector<pair<int, int>> freqVec(freqMap.begin(), freqMap.end());
sort(freqVec.begin(), freqVec.end(), [](pair<int, int> &p1, pair<int, int> &p2) {
return p1.second > p2.second;
});
for (auto freqPair: freqVec) {
cout << freqPair.first << ",";
}
cout << "\b" << endl;
return 0;
}
Node.js版本
const rl = require("readline").createInterface({ input: process.stdin });
var iter = rl[Symbol.asyncIterator]();
const readline = async () => (await iter.next()).value;
void async function () {
let numbers = (await readline()).split(',').map(i => parseInt(i))
const freqMap = new Map()
for (let number of numbers) {
if (freqMap.has(number)) {
freqMap.set(number, freqMap.get(number) + 1)
} else {
freqMap.set(number, 1)
}
}
const freqArray = Array.from(freqMap.entries());
freqArray.sort((a, b) => {
return b[1] - a[1];
});
const result = freqArray.map(entry => entry[0]).join(',');
console.log(result);
}()
C语言版本
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct number {
int val;
int cnt;
} Number;
int cmp(const void *a, const void *b) {
if (((Number *) a)->cnt != ((Number *) b)->cnt) {
return ((Number *) b)->cnt - ((Number *) a)->cnt;
}
return ((Number *) a)->val - ((Number *) b)->val;
}
int main() {
setbuf(stdout, NULL);
char str[1024] = {0};
gets(str);
char *p = str;
Number arr[1024] = {0};
int arrLen = 0;
while (p = strtok(p, ",")) {
int flag = 1;
for (int i = 0; i < arrLen; ++i) {
if (arr[i].val == atoi(p)) {
flag = 0;
arr[i].cnt++;
break;
}
}
if (flag == 1) {
arr[arrLen].val = atoi(p);
arr[arrLen].cnt = 1;
arrLen++;
}
p = NULL;
}
qsort(arr, arrLen, sizeof(arr[0]), cmp);
for (int i = 0; i < arrLen; ++i) {
printf("%d,", arr[i].val);
}
printf("\b");
return 0;
}
文章来源:https://blog.csdn.net/qq_43662348/article/details/135260454
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