排序数组中的最后一个重复元素
原文： https://www.geeksforgeeks.org/last-duplicate-element-sorted-array/
我们有一个包含重复元素的排序数组，我们必须找到最后一个重复元素的索引并打印它的索引，还要打印重复元素。 如果找不到这样的元素，则打印一条消息。
示例：
Input : arr[] = {1, 5, 5, 6, 6, 7}
Output :
Last index: 4
Last duplicate item: 6
Input : arr[] = {1, 2, 3, 4, 5}
Output : No duplicate found
我们只需以相反的顺序遍历数组，然后比较当前元素和上一个元素。 如果找到匹配项，则打印索引和重复元素。 由于这是排序数组，它将是最后一个重复项。 如果找不到此类元素，我们将为其打印消息。
1- for i = n-1 to 0
if (arr[i] == arr[i-1])
Print current element and its index.
Return
2- If no such element found print a message
of no duplicate found.
C++
// To print last duplicate element and its
// index in a sorted array
#include <bits/stdc++.h>
void dupLastIndex(int arr[], int n) {
// if array is null or size is less
// than equal to 0 return
if (arr == NULL || n <= 0)
return;
// compare elements and return last
// duplicate and its index
for (int i = n - 1; i > 0; i--) {
if (arr[i] == arr[i - 1]) {
printf("Last index: %d\nLast "
"duplicate item: %d\n", i, arr[i]);
return;
}
}
// If we reach here, then no duplicate
// found.
printf("no duplicate found");
}
int main() {
int arr[] = {1, 5, 5, 6, 6, 7, 9};
int n = sizeof(arr) / sizeof(int);
dupLastIndex(arr, n);
return 0;
}
Java
// Java code to print last duplicate element
// and its index in a sorted array
import java.io.*;
class GFG
{
static void dupLastIndex(int arr[], int n)
{
// if array is null or size is less
// than equal to 0 return
if (arr == null || n <= 0)
return;
// compare elements and return last
// duplicate and its index
for (int i = n - 1; i > 0; i--)
{
if (arr[i] == arr[i - 1])
{
System.out.println("Last index:" + i);
System.out.println("Last duplicate item: "
+ arr[i]);
return;
}
}
// If we reach here, then no duplicate
// found.
System.out.print("no duplicate found");
}
// Driver code
public static void main (String[] args)
{
int arr[] = {1, 5, 5, 6, 6, 7, 9};
int n = arr.length;
dupLastIndex(arr, n);
}
}
// This code is contributed by vt_m
Python3
# Python3 code to print last duplicate
# element and its index in a sorted array
def dupLastIndex(arr, n):
# if array is null or size is less
# than equal to 0 return
if (arr == None or n <= 0):
return
# compare elements and return last
# duplicate and its index
for i in range(n - 1, 0, -1):
if (arr[i] == arr[i - 1]):
print("Last index:", i, "\nLast",
"duplicate item:",arr[i])
return
# If we reach here, then no duplicate
# found.
print("no duplicate found")
arr = [1, 5, 5, 6, 6, 7, 9]
n = len(arr)
dupLastIndex(arr, n)
# This code is contributed by
# Smitha Dinesh Semwal
C#
// C# code to print last duplicate element
// and its index in a sorted array
using System;
class GFG {
static void dupLastIndex(int []arr, int n)
{
// if array is null or size is less
// than equal to 0 return
if (arr == null || n <= 0)
return;
// compare elements and return last
// duplicate and its index
for (int i = n - 1; i > 0; i--)
{
if (arr[i] == arr[i - 1])
{
Console.WriteLine("Last index:" + i);
Console.WriteLine("Last duplicate item: "
+ arr[i]);
return;
}
}
// If we reach here, then no duplicate
// found.
Console.WriteLine("no duplicate found");
}
// Driver code
public static void Main ()
{
int []arr = {1, 5, 5, 6, 6, 7, 9};
int n = arr.Length;
dupLastIndex(arr, n);
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP program to print last
// duplicate element and its
// index in a sorted array
function dupLastIndex($arr, $n)
{
// if array is null or size is less
// than equal to 0 return
if ($arr == null or $n <= 0)
return;
// compare elements and return last
// duplicate and its index
for ( $i = $n - 1; $i > 0; $i--)
{
if ($arr[$i] == $arr[$i - 1])
{
echo "Last index:", $i , "\n";
echo "Last duplicate item:", $arr[$i];
return;
}
}
// If we reach here, then
// no duplicate found.
echo "no duplicate found";
}
// Driver Code
$arr = array(1, 5, 5, 6, 6, 7, 9);
$n = count($arr);
dupLastIndex($arr, $n);
// This code is contributed by anuj_67\.
?>
输出：
Last index: 4
Last duplicate item: 6
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