存储重复的键值对，并按键对键值对进行排序

给定N个包含重复键和值的键/值对，任务是存储这些对并按键对对进行排序。

例子：

Input : 
N : 10 
Keys : 5 1 4 6 8 0 6 6 5 5 
values: 0 1 2 3 4 5 6 7 8 9 
Output : 
Keys : 0 1 4 5 5 5 6 6 6 8 
values: 5 1 2 0 8 9 3 6 7 4 
Explanation: 
We have given 10 key, value pairs which contain duplicate keys and values. 
The key value pair is {(5, 0), (1, 1), (4, 2), (6, 3), (8, 4), (0, 5), (6, 6), 
(6, 7), (5, 8), (5, 9)} and we want to store these key value pair and sort these 
key value pair by keys. So, the expected output is {(0, 5), (1, 1), (4, 2), (5, 0), 
(5, 8), (5, 9), (6, 3), (6, 6), (6, 7), (8, 4)}. Because the sorted 
increasing order of keys is {0, 1, 4, 5, 5, 5, 6, 6, 6, 8}. 
 

方法：
为了解决上述问题，我们可以使用单独的数组存储键和值，然后可以简单地通过Merge排序算法对键进行排序。我们可以使用任何排序算法，但Mergesort是最快的标准排序算法，并且并行地，我们可以对在键上执行的values数组执行相同的操作，以便键值对将在两个数组上保持相同的索引。

下面是上述方法的实现：

Java

// Java program to Store duplicate
// keys-values pair and sort the
// key-value pair by key
import java.util.*;
import java.lang.*;
import java.io.*;
 
class Solution {
 
    // Merges two subarrays of arr[].
    // First subarray is arr[l..m]
    // Second subarray is arr[m+1..r]
    void merge(int arrk[], int l, int m,
            int r, int arrv[])
    {
        // Sizes of two subarrays
        // that are to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temporary arrays */
        int L[] = new int[n1];
        int R[] = new int[n2];
        int Lk[] = new int[n1];
        int Rk[] = new int[n2];
 
        /* Copy data to temporary arrays */
        for (int i = 0; i < n1; ++i) {
            L[i] = arrk[l + i];
            Lk[i] = arrv[l + i];
        }
        for (int j = 0; j < n2; ++j) {
            R[j] = arrk[m + 1 + j];
            Rk[j] = arrv[m + 1 + j];
        }
 
        // Initial indexes of
        // first and second subarrays
        int i = 0, j = 0;
 
        int k = l;
 
        while (i < n1 && j < n2) {
            if (L[i] <= R[j]) {
                arrk[k] = L[i];
                arrv[k] = Lk[i];
 
                i++;
            }
            else {
                arrk[k] = R[j];
                arrv[k] = Rk[j];
 
                j++;
            }
 
            k++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1) {
            arrk[k] = L[i];
            arrv[k] = Lk[i];
 
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2) {
            arrk[k] = R[j];
            arrv[k] = Rk[j];
 
            j++;
            k++;
        }
    }
 
    // Function that sorts arr[l..r] using merge()
    void sort(int arrk[], int l, int r, int arrv[])
    {
        if (l < r) {
            // Find the middle point
            int m = (l + r) / 2;
 
            // Sort first and second halves
            sort(arrk, l, m, arrv);
            sort(arrk, m + 1, r, arrv);
 
            // Merge the sorted halves
            merge(arrk, l, m, r, arrv);
        }
    }
 
    /* Function to print array of size n */
    static void printArray(int arr[])
    {
        int n = arr.length;
 
        for (int i = 0; i < n; ++i)
 
            System.out.print(arr[i] + " ");
 
        System.out.println();
    }
 
    // Driver code
    public static void main(String[] args)
                    throws java.lang.Exception
    {
 
        // Size of Array
        int n = 10;
 
        // array of keys
        int[] arrk = { 5, 1, 4, 6, 8, 0,
                                    6, 6, 5, 5 };
 
        // array of values
        int[] arrv = { 0, 1, 2, 3, 4, 5,
                                    6, 7, 8, 9 };
 
        Solution ob = new Solution();
 
        ob.sort(arrk, 0, n - 1, arrv);
 
        System.out.print("Keys: ");
        printArray(arrk);
 
        System.out.println();
 
        System.out.print("Values: ");
        printArray(arrv);
    }
}

C#

// C# program to Store duplicate
// keys-values pair and sort the
// key-value pair by key
using System;
 
class Solution{
 
// Merges two subarrays of arr[].
// First subarray is arr[l..m]
// Second subarray is arr[m+1..r]
void merge(int []arrk, int l, int m,
           int r, int []arrv)
{
     
    // Sizes of two subarrays
    // that are to be merged
    int n1 = m - l + 1;
    int n2 = r - m;
 
    // Create temporary arrays
    int []L = new int[n1];
    int []R = new int[n2];
    int []Lk = new int[n1];
    int []Rk = new int[n2];
 
    // Copy data to temporary arrays
    for(int i = 0; i < n1; ++i)
    {
        L[i] = arrk[l + i];
        Lk[i] = arrv[l + i];
    }
    for(int j = 0; j < n2; ++j)
    {
        R[j] = arrk[m + 1 + j];
        Rk[j] = arrv[m + 1 + j];
    }
 
    // Initial indexes of
    // first and second subarrays
    int a = 0 , b = 0;
 
    int k = l;
 
    while (a < n1 && b < n2)
    {
        if (L[a] <= R[b])
        {
            arrk[k] = L[a];
            arrv[k] = Lk[a];
 
            a++;
        }
        else
        {
            arrk[k] = R[b];
            arrv[k] = Rk[b];
 
            b++;
        }
        k++;
    }
 
    // Copy remaining elements of L[] if any
    while (a < n1)
    {
        arrk[k] = L[a];
        arrv[k] = Lk[a];
 
        a++;
        k++;
    }
 
    // Copy remaining elements of R[] if any
    while (b < n2)
    {
        arrk[k] = R[b];
        arrv[k] = Rk[b];
 
        b++;
        k++;
    }
}
 
// Function that sorts arr[l..r] using merge()
void sort(int []arrk, int l, int r, int []arrv)
{
    if (l < r)
    {
         
        // Find the middle point
        int m = (l + r) / 2;
 
        // Sort first and second halves
        sort(arrk, l, m, arrv);
        sort(arrk, m + 1, r, arrv);
 
        // Merge the sorted halves
        merge(arrk, l, m, r, arrv);
    }
}
 
// Function to print array of size n
static void printArray(int []arr)
{
    int n = arr.Length;
 
    for(int i = 0; i < n; ++i)
        Console.Write(arr[i] + " ");
 
    Console.WriteLine();
}
 
// Driver code
public static void Main(string[] args)
{
     
    // Size of Array
    int n = 10;
 
    // array of keys
    int[] arrk = { 5, 1, 4, 6, 8,
                   0, 6, 6, 5, 5 };
 
    // array of values
    int[] arrv = { 0, 1, 2, 3, 4,
                   5, 6, 7, 8, 9 };
 
    Solution ob = new Solution();
 
    ob.sort(arrk, 0, n - 1, arrv);
 
    Console.Write("Keys: ");
    printArray(arrk);
 
    Console.WriteLine();
 
    Console.Write("Values: ");
    printArray(arrv);
}
}
 
// This code is contributed by ukasp

Javascript

 
// Javascript program to Store duplicate
// keys-values pair and sort the
// key-value pair by key
 
// Merges two subarrays of arr[].
// First subarray is arr[l..m]
// Second subarray is arr[m+1..r]
function merge(arrk, l, m, r, arrv)
{
    // Sizes of two subarrays
    // that are to be merged
    var n1 = m - l + 1;
    var n2 = r - m;
     
    /* Create temporary arrays */
    var L = Array(n1).fill(0);
    var R = Array(n2).fill(0);
    var Lk = Array(n1).fill(0);
    var Rk = Array(n2).fill(0);
     
    /* Copy data to temporary arrays */
    for (var i = 0; i < n1; ++i) {
        L[i] = arrk[l + i];
        Lk[i] = arrv[l + i];
    }
    for (var j = 0; j < n2; ++j) {
        R[j] = arrk[m + 1 + j];
        Rk[j] = arrv[m + 1 + j];
    }
    // Initial indexes of
    // first and second subarrays
    var i = 0, j = 0;
    var k = l;
    while (i < n1 && j < n2) {
        if (L[i] <= R[j]) {
            arrk[k] = L[i];
            arrv[k] = Lk[i];
            i++;
        }
        else {
            arrk[k] = R[j];
            arrv[k] = Rk[j];
            j++;
        }
        k++;
    }
    /* Copy remaining elements of L[] if any */
    while (i < n1) {
        arrk[k] = L[i];
        arrv[k] = Lk[i];
        i++;
        k++;
    }
    /* Copy remaining elements of R[] if any */
    while (j < n2) {
        arrk[k] = R[j];
        arrv[k] = Rk[j];
        j++;
        k++;
    }
}
// Function that sorts arr[l..r] using merge()
function sort(arrk, l, r, arrv)
{
    if (l < r) {
        // Find the middle point
        var m = parseInt((l + r) / 2);
        // Sort first and second halves
        sort(arrk, l, m, arrv);
        sort(arrk, m + 1, r, arrv);
        // Merge the sorted halves
        merge(arrk, l, m, r, arrv);
    }
}
/* Function to print array of size n */
function printArray(arr)
{
    var n = arr.length;
    for (var i = 0; i < n; ++i)
        document.write(arr[i] + " ");
    document.write("<br>");
}
// Driver code
 
// Size of Array
var n = 10;
// array of keys
var arrk = [ 5, 1, 4, 6, 8, 0, 6, 6, 5, 5 ]
// array of values
var arrv = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ]
sort(arrk, 0, n - 1, arrv);
document.write("Keys: ");
printArray(arrk);
document.write("<br>");
document.write("Values: ");
printArray(arrv);
 

输出：

Keys: 0 1 4 5 5 5 6 6 6 8 

Values: 5 1 2 0 8 9 3 6 7 4

时间复杂度： O(N * logN)