华为OD机试新系统真题【终端设备档位差异统计】

终端设备档位差异统计(C/C++/Py/Java/Js/Go)题解

华为OD机试新系统真题 华为OD上机考试新系统真题 6月24号 100分题型

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题目内容

在进行手机、平板、穿戴等设备的系统调优时，会为不同设备采集一组性能与能耗综合档位值；为了分析产品矩阵中是否存在稳定的档位梯度，工程师需要统计所有 “档位差异恰好为指定值” 的档位组合数量。 给定一个整数数组

p

r

o

f

i

l

e

s

profiles

profiles，其中

p

r

o

f

i

l

e

s

[

i

]

profiles[i]

profiles[i] 表示第

i

i

i 台终端设备的综合档位值；再给定一个非负整数

d

i

f

f

diff

diff，表示目标档位差。 请你返回不同档位值组成的组合数量，使得两个档位值的绝对差等于

d

i

f

f

diff

diff。

注意：

约束条件：

• $1\le$
• $-2147483648\le$
• $0\le diff\le 2147483647$

样例1

输入

1,5,3,4,2
2

输出

3

说明 满足要求的不同档位组合为

(

1

,

3

)

(1, 3)

(1,3)、

(

3

,

5

)

(3, 5)

(3,5)、

(

2

,

4

)

(2, 4)

(2,4)，共

3

3

3 组。

样例2

输入

8,8,8,10,10,12
0

输出

2

说明 档位

8

8

8 和档位

10

10

10 都至少出现两次，可以分别形成一组差值为

0

0

0 的组合；档位

12

12

12 只出现一次，不能形成组合。

样例3

输入

-5,-3,-1,1
2

输出

3

说明 满足要求的组合为

(

−

5

,

−

3

)

(-5, -3)

(−5,−3)、

(

−

3

,

−

1

)

(-3, -1)

(−3,−1)、

(

−

1

,

1

)

(-1, 1)

(−1,1)。

题解

思路：逻辑分析

C++

#include<bits/stdc++.h>
#include <unordered_map>
#include <vector>
using namespace std;

// 通用 切割函数 函数 将字符串str根据delimiter进行切割
vector<int> split(const string& str, const string& delimiter) {
vector<int> result;
size_t start = 0;
size_t end = str.find(delimiter);
while (end != string::npos) {
result.push_back(stoi(str.substr(start, end – start)));
start = end + delimiter.length();
end = str.find(delimiter, start);
}
// 添加最后一个部分
result.push_back(stoi(str.substr(start)));
return result;
}

int countProfileDiff(vector<int>& profiles, int diff) {
if (diff == 0) {
unordered_map<long, int> numCount;
for (auto& profile : profiles) {
numCount[profile]++;
}
int res = 0;
// 统计次数超过1次的数量
for (auto& p : numCount) {
if (p.second > 1) {
res++;
}
}
return res;
}
int res = 0;

unordered_set<long> s(profiles.begin(), profiles.end());
for (auto num : s) {
if (s.count(num + diff)) {
res++;
}
}
return res;
}

int main() {
string input;
int diff;
getline(cin, input);
cin >> diff;
vector<int> profiles = split(input, ",");
cout << countProfileDiff(profiles, diff);
return 0;
}

Java

import java.io.*;
import java.util.*;

public class Main {

static int countProfileDiff(int[] profiles, int diff) {

if (diff == 0) {
Map<Long, Integer> numCount = new HashMap<>();

for (int profile : profiles) {
numCount.put((long) profile,
numCount.getOrDefault((long) profile, 0) + 1);
}

int res = 0;

// 统计次数超过1次的数量
for (int cnt : numCount.values()) {
if (cnt > 1) {
res++;
}
}

return res;
}

int res = 0;

Set<Long> set = new HashSet<>();

for (int profile : profiles) {
set.add((long) profile);
}

for (long num : set) {
if (set.contains(num + diff)) {
res++;
}
}

return res;
}

public static void main(String[] args) throws Exception {

BufferedReader br = new BufferedReader(
new InputStreamReader(System.in));

String input = br.readLine();
int diff = Integer.parseInt(br.readLine());

String[] parts = input.split(",");

int[] profiles = new int[parts.length];

for (int i = 0; i < parts.length; i++) {
profiles[i] = Integer.parseInt(parts[i]);
}

System.out.println(countProfileDiff(profiles, diff));
}
}

Python

from collections import Counter

def count_profile_diff(profiles, diff):

if diff == 0:
num_count = Counter(profiles)

res = 0

# 统计次数超过1次的数量
for cnt in num_count.values():
if cnt > 1:
res += 1

return res

res = 0

s = set(profiles)

for num in s:
if num + diff in s:
res += 1

return res

profiles = list(map(int, input().split(",")))
diff = int(input())

print(count_profile_diff(profiles, diff))

JavaScript

const readline = require("readline");

const rl = readline.createInterface({
input: process.stdin,
output: process.stdout
});

const lines = [];

rl.on("line", line => {
lines.push(line);
});

rl.on("close", () => {

const profiles = lines[0].split(",").map(Number);
const diff = Number(lines[1]);

function countProfileDiff(profiles, diff) {

if (diff === 0) {

const numCount = new Map();

for (const profile of profiles) {
numCount.set(
profile,
(numCount.get(profile) || 0) + 1
);
}

let res = 0;

// 统计次数超过1次的数量
for (const cnt of numCount.values()) {
if (cnt > 1) {
res++;
}
}

return res;
}

let res = 0;

const s = new Set(profiles);

for (const num of s) {
if (s.has(num + diff)) {
res++;
}
}

return res;
}

console.log(countProfileDiff(profiles, diff));
});

Go

package main

import (
"bufio"
"fmt"
"os"
"strconv"
"strings"
)

func countProfileDiff(profiles []int, diff int) int {

if diff == 0 {

numCount := make(map[int]int)

for _, profile := range profiles {
numCount[profile]++
}

res := 0

// 统计次数超过1次的数量
for _, cnt := range numCount {
if cnt > 1 {
res++
}
}

return res
}

res := 0

s := make(map[int]bool)

for _, profile := range profiles {
s[profile] = true
}

for num := range s {
if s[num+diff] {
res++
}
}

return res
}

func main() {

reader := bufio.NewReader(os.Stdin)

line, _ := reader.ReadString('\\n')
line = strings.TrimSpace(line)

diffLine, _ := reader.ReadString('\\n')
diffLine = strings.TrimSpace(diffLine)

diff, _ := strconv.Atoi(diffLine)

parts := strings.Split(line, ",")

profiles := make([]int, len(parts))

for i, part := range parts {
profiles[i], _ = strconv.Atoi(part)
}

fmt.Println(countProfileDiff(profiles, diff))
}

C语言

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAXN 100000

// qsort比较函数（升序）
int cmp(const void *a, const void *b) {
int x = *(const int *)a;
int y = *(const int *)b;

if (x < y) return –1;
if (x > y) return 1;
return 0;
}

int countProfileDiff(int profiles[], int n, int diff) {
// 先升序排序
qsort(profiles, n, sizeof(int), cmp);

int res = 0;
long nums[MAXN];
// 已经访问过去重的数量
int len = 0;
// 当前遍历元素数量
int index = 0;
for (int i = 0; i < n; i++) {
long val = (long)profiles[i] – (long)diff;
while (index < len && nums[index] < val) {
index++;
}

if (index < len && nums[index] == val) {
res++;
// 只能进行一次组合
index++;
}
// 将当前元素添加到nums中，注意去重
if ((len == 0 || nums[len – 1] != (long) profiles[i])) {
nums[len++] = (long) profiles[i];
}
}

return res;
}

int main() {

char input[10000000];

fgets(input, sizeof(input), stdin);

input[strcspn(input, "\\n")] = '\\0';

int diff;
scanf("%d", &diff);

int profiles[MAXN];
int n = 0;

char *token = strtok(input, ",");

while (token != NULL) {
profiles[n++] = atoi(token);
token = strtok(NULL, ",");
}

printf("%d\\n", countProfileDiff(profiles, n, diff));

return 0;
}