按相对顺序打印数组中最后出现的元素
给定一个由 N 个元素组成的数组，按照给定的相对顺序打印元素，删除除最后一次出现的元素之外的所有出现的元素。 例 :
输入: a[] = {1，5，5，1，6，1} 输出: 5 6 1 去掉两个整数 1，在位置 1 和 4。另外，删除整数 5，它位于位置 2。 因此左数组为{5，6，1} 输入: a[] = {2，5，5，2} 输出: 5 2
进场:
- 散列每个元素的最后一次出现。
- 迭代 N 个元素的数组，如果元素的索引被散列，那么打印数组元素。
以下是上述方法的实现:
C++
// C++ program to print the last occurrence
// of every element in relative order
#include <bits/stdc++.h>
using namespace std;
// Function to print the last occurrence
// of every element in an array
void printLastOccurrence(int a[], int n)
{
// used in hashing
unordered_map<int, int> mp;
// iterate and store the last index
// of every element
for (int i = 0; i < n; i++)
mp[a[i]] = i;
// iterate and check for the last
// occurrence of every element
for (int i = 0; i < n; i++) {
if (mp[a[i]] == i)
cout << a[i] << " ";
}
}
// Driver Code
int main()
{
int a[] = { 1, 5, 5, 1, 6, 1 };
int n = sizeof(a) / sizeof(a[0]);
printLastOccurrence(a, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to print the
// last occurrence of every
// element in relative order
import java.util.*;
class GFG
{
// Function to print the last
// occurrence of every element
// in an array
public static void printLastOccurrence(int a[],
int n)
{
HashMap<Integer,
Integer> map = new HashMap<Integer,
Integer>();
// iterate and store the last
// index of every element
for (int i = 0; i < n; i++)
map.put(a[i], i);
for (int i = 0; i < n; i++)
{
if (map.get(a[i]) == i)
System.out.print(a[i] +" ");
}
}
// Driver Code
public static void main (String[] args)
{
int a[] = { 1, 5, 5, 1, 6, 1 };
int n = a.length;
printLastOccurrence(a, n);
}
}
// This code is contributed
// by ankita_saini
Python 3
# Python 3 program to print the last occurrence
# of every element in relative order
# Function to print the last occurrence
# of every element in an array
def printLastOccurrence(a, n):
# used in hashing
mp = {i:0 for i in range(7)}
# iterate and store the last
# index of every element
for i in range(n):
mp[a[i]] = i
# iterate and check for the last
# occurrence of every element
for i in range(n):
if (mp[a[i]] == i):
print(a[i], end = " ")
# Driver Code
if __name__ == '__main__':
a = [1, 5, 5, 1, 6, 1]
n = len(a)
printLastOccurrence(a, n)
# This code is contributed by
# Surendra_Gangwar
C
// C# program to print the
// last occurrence of every
// element in relative order
using System;
class GFG
{
// Function to print the last
// occurrence of every element
// in an array
public static void printLastOccurrence(int[] a,
int n)
{
HashMap<Integer,
Integer> map = new HashMap<Integer,
Integer>();
// iterate and store the last
// index of every element
for (int i = 0; i < n; i++)
map.put(a[i], i);
for (int i = 0; i < n; i++)
{
if (map.get(a[i]) == i)
Console.Write(a[i] + " ");
}
}
// Driver Code
public static void Main ()
{
int[] a = { 1, 5, 5, 1, 6, 1 };
int n = a.Length;
printLastOccurrence(a, n);
}
}
// This code is contributed
// by ChitraNayal
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to print the last
// occurrence of every element
// in relative order
// Function to print the last
// occurrence of every element
// in an array
function printLastOccurrence(&$a, $n)
{
// used in hashing
$mp = array();
// iterate and store the last
// index of every element
for ($i = 0; $i < $n; $i++)
$mp[$a[$i]] = $i;
// iterate and check for the last
// occurrence of every element
for ($i = 0; $i < $n; $i++)
{
if ($mp[$a[$i]] == $i)
echo $a[$i] . " ";
}
}
// Driver Code
$a = array(1, 5, 5, 1, 6, 1);
$n = sizeof($a);
printLastOccurrence($a, $n);
// This code is contributed
// by ChitraNayal
?>
java 描述语言
<script>
// Javascript program to print the last
// occurrence of every element
// in relative order
// Function to print the last
// occurrence of every element
// in an array
function printLastOccurrence(a, n)
{
// used in hashing
let mp = [];
// iterate and store the last
// index of every element
for (let i = 0; i < n; i++)
mp[a[i]] = i;
// iterate and check for the last
// occurrence of every element
for (let i = 0; i < n; i++)
{
if (mp[a[i]] == i)
document.write( a[i] + " ");
}
}
// Driver Code
let a = [1, 5, 5, 1, 6, 1];
let n = a.length;
printLastOccurrence(a, n);
// This code is contributed by sravan kumar
</script>
输出:
5 6 1
时间复杂度: O(N * log N)
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