原文:
给定一个由 n 个不同整数组成的 数组【arr】，任务是为每个数组元素打印其左侧所有个更大的元素。
示例:
输入: arr[] = {5，3，9，0，16，12} 输出: 5: 3:5 9: 0:9 5 3 16: 12:16
输入: arr[] = {1，2，0} 输出: 1: 2: 0: 2 1
天真方法:解决这个问题最简单的方法是遍历数组，对于每个数组元素，遍历它前面的所有元素，打印大于当前元素的元素。
c
// c program for naive approach
#include
using namespace std;
// function to print all greater elements on the left of each array element
void printgreater(vector& arr)
{
// store the size of array in variable n
int n = arr.size();
for (int i = 0; i < n; i )
{
// result is used to store elements which are greater than current element present left side of current element
vector result;
// traversing all the elements which are left of current element arr[i]
for (int j = i - 1; j >= 0; j--)
{
//checking whether arr[j] (left element) is greater than current element or not
// if yes then insert the element to the result vector
if (arr[j] > arr[i])
{
result.push_back(arr[j]);
}
}
cout << arr[i] << ": ";
//printing all the elements present in result vector
for (int k = 0; k < result.size(); k )
{
cout << result[k] << " ";
}
cout << endl;
}
}
//driver code
int main()
{
vector arr{5, 3, 9, 0, 16, 12};
printgreater(arr);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program for naive approach
import java.util.*;
class gfg{
// function to print all greater elements on the left of each array element
static void printgreater(arraylist arr)
{
// store the size of array in variable n
int n = arr.size();
for (int i = 0; i < n; i )
{
// result is used to store elements which
// are greater than current element present
// left side of current element
arraylist result
= new arraylist();
// traversing all the elements which
// are left of current element arr[i]
for (int j = i - 1; j >= 0; j--)
{
// checking whether arr[j] (left element) is
// greater than current element or not
// if yes then insert the element to the result vector
if (arr.get(j) > arr.get(i))
{
result.add(arr.get(j));
}
}
system.out.print(arr.get(i) ": ");
// printing all the elements present in result vector
for (int k = 0; k < result.size(); k )
{
system.out.print(result.get(k) " ");
}
system.out.println();
}
}
// driver code
public static void main(string args[])
{
arraylist arr = new arraylist(arrays.aslist(5, 3, 9, 0, 16, 12));
printgreater(arr);
}
}
// this code is contributed by bgangwar59.
python 3
# python3 program for naive approach
# function to print all greater
# elements on the left of each
# array element
def printgreater(arr):
# store the size of array
# in variable n
n = len(arr)
for i in range(n):
# result is used to store elements
# which are greater than current
# element present left side of
# current element
result = []
# traversing all the elements
# which are left of current
# element arr[i]
j = i - 1
while(j >= 0):
# checking whether arr[j] (left element)
# is greater than current element or not
# if yes then insert the element to the
# result vector
if (arr[j] > arr[i]):
result.append(arr[j])
j -= 1
print(arr[i], end = ": ")
# printing all the elements present
# in result vector
for k in range(len(result)):
print(result[k], end = " ")
print("\n", end = "")
# driver code
if __name__ == '__main__':
arr = [ 5, 3, 9, 0, 16, 12 ]
printgreater(arr)
# this code is contributed by ipg2016107
c
// c# program for naive approach
using system;
using system.collections.generic;
class gfg
{
// function to print all greater elements on the left of
// each array element
static void printgreater(int[] arr)
{
// store the size of array in variable n
int n = arr.length;
for (int i = 0; i < n; i )
{
// result is used to store elements which are
// greater than current element present left
// side of current element
list result = new list();
// traversing all the elements which are left of
// current element arr[i]
for (int j = i - 1; j >= 0; j--)
{
// checking whether arr[j] (left element) is
// greater than current element or not
// if yes then insert the element to the
// result vector
if (arr[j] > arr[i]) {
result.add(arr[j]);
}
}
console.write(arr[i] ": ");
// printing all the elements present in result
// vector
for (int k = 0; k < result.count; k ) {
console.write(result[k] " ");
}
console.writeline();
}
}
// driver code
public static void main()
{
int[] arr = { 5, 3, 9, 0, 16, 12 };
printgreater(arr);
}
}
// this code is contributed by ukasp.
java 描述语言
output
5:
3: 5
9:
0: 9 3 5
16:
12: 16
时间复杂度:o(n2) t5】辅助空间: o(n)
高效方法:为了优化上述方法，想法是利用一个。是使用自平衡 bst 实现的，可以用来解决这个问题。按照以下步骤解决问题:
- 初始化一个和集，以非递减顺序存储元素。
- 在索引 0 到n–1上遍历数组，并执行以下操作:
- t4 中。
- 开始迭代到 p 并打印集合中的元素。
以下是上述方法的实现:
c
// c program for the above approach
#include
using namespace std;
// function to print all greater elements
// on the left of each array element
void printgreater(vector& arr)
{
int n = arr.size();
// set to implement
// self-balancing bsts
set > s;
// traverse the array
for (int i = 0; i < n; i ) {
// insert the current
// element into the set
auto p = s.insert(arr[i]);
auto j = s.begin();
cout << arr[i] << ": ";
// iterate through the set
while (j != p.first) {
// print the element
cout << *j << " ";
j ;
}
cout << endl;
}
}
// driver code
int main()
{
vector arr{ 5, 3, 9, 0, 16, 12 };
printgreater(arr);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class gfg{
// function to print all greater elements
// on the left of each array element
static void printgreater(int arr[])
{
int n = arr.length;
// set to implement
// self-balancing bsts
treeset s = new treeset<>(
collections.reverseorder());
// traverse the array
for(int i = 0; i < n; i )
{
// insert the current
// element into the set
s.add(arr[i]);
system.out.print(arr[i] ": ");
// iterate through the set
for(int v : s)
{
if (v == arr[i])
break;
// print the element
system.out.print(v " ");
}
system.out.println();
}
}
// driver code
public static void main(string[] args)
{
int arr[] = { 5, 3, 9, 0, 16, 12 };
printgreater(arr);
}
}
// this code is contributed by kingash
python 3
# python3 program for the above approach
# function to print all greater elements
# on the left of each array element
def printgreater(arr):
n = len(arr)
# set to implement
# self-balancing bsts
s = set([])
# traverse the array
for i in range(n):
# insert the current
# element into the set
s.add(arr[i])
print(arr[i], ": ", sep = "", end = "")
temp = list(s)
temp.sort()
temp.reverse()
# iterate through the set
for v in range(len(temp)):
if (temp[v] == arr[i]):
break
# print the element
print(temp[v], end = " ")
print()
arr = [5, 3, 9, 0, 16, 12]
printgreater(arr)
# this code is contributed by divyesh072019.
c
// c# program for the above approach
using system;
using system.collections.generic;
class gfg {
// function to print all greater elements
// on the left of each array element
static void printgreater(int[] arr)
{
int n = arr.length;
// set to implement
// self-balancing bsts
hashset s = new hashset();
// traverse the array
for(int i = 0; i < n; i )
{
// insert the current
// element into the set
s.add(arr[i]);
console.write(arr[i] ": ");
list temp = new list();
// iterate through the set
foreach(int v in s)
{
temp.add(v);
}
temp.sort();
temp.reverse();
// iterate through the set
for(int v = 0; v < temp.count; v )
{
if (temp[v] == arr[i])
{
break;
}
// print the element
console.write(temp[v] " ");
}
console.writeline();
}
}
// driver code
static void main() {
int[] arr = { 5, 3, 9, 0, 16, 12 };
printgreater(arr);
}
}
// this code is contributed by rameshtravel07.
java 描述语言
output
5:
3: 5
9:
0: 9 5 3
16:
12: 16
时间复杂度:o(n2) 辅助空间: o(n)
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