打印给定范围内所有数字严格递增的数字
原文:https://www . geesforgeks . org/print-给定范围内的所有数字都有严格递增顺序的数字/
给定两个正整数 L 和 R ，任务是打印【L，R】范围内的数字，这些数字的位数严格递增。
示例:
输入: L = 10，R = 15 输出: 12 13 14 15 说明: 在[10，15]范围内，只有数字{12，13，14，15}的位数是严格递增的。
输入: L = 60，R = 70 输出: 67 68 69 说明: 在【60，70】范围内，只有数字{67，68，69}的位数是严格递增的。
方法:思路是迭代范围【L，R】，对于该范围内的每个数字，检查该数字的位数是否严格递增。如果是，则打印该号码，否则检查下一个号码。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print all numbers
// in the range [L, R] having digits
// in strictly increasing order
void printNum(int L, int R)
{
// Iterate over the range
for (int i = L; i <= R; i++) {
int temp = i;
int c = 10;
int flag = 0;
// Iterate over the digits
while (temp > 0) {
// Check if the current digit
// is >= the previous digit
if (temp % 10 >= c) {
flag = 1;
break;
}
c = temp % 10;
temp /= 10;
}
// If the digits are in
// ascending order
if (flag == 0)
cout << i << " ";
}
}
// Driver Code
int main()
{
// Given range L and R
int L = 10, R = 15;
// Function Call
printNum(L, R);
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.util.*;
class GFG{
// Function to print all numbers
// in the range [L, R] having digits
// in strictly increasing order
static void printNum(int L, int R)
{
// Iterate over the range
for(int i = L; i <= R; i++)
{
int temp = i;
int c = 10;
int flag = 0;
// Iterate over the digits
while (temp > 0)
{
// Check if the current digit
// is >= the previous digit
if (temp % 10 >= c)
{
flag = 1;
break;
}
c = temp % 10;
temp /= 10;
}
// If the digits are in
// ascending order
if (flag == 0)
System.out.print(i + " ");
}
}
// Driver code
public static void main(String[] args)
{
// Given range L and R
int L = 10, R = 15;
// Function call
printNum(L, R);
}
}
// This code is contributed by offbeat
Python 3
# Python3 program for the above approach
# Function to print all numbers in
# the range [L, R] having digits
# in strictly increasing order
def printNum(L, R):
# Iterate over the range
for i in range(L, R + 1):
temp = i
c = 10
flag = 0
# Iterate over the digits
while (temp > 0):
# Check if the current digit
# is >= the previous digit
if (temp % 10 >= c):
flag = 1
break
c = temp % 10
temp //= 10
# If the digits are in
# ascending order
if (flag == 0):
print(i, end = " ")
# Driver Code
# Given range L and R
L = 10
R = 15
# Function call
printNum(L, R)
# This code is contributed by code_hunt
C
// C# program for the above approach
using System;
class GFG{
// Function to print all numbers
// in the range [L, R] having digits
// in strictly increasing order
static void printNum(int L, int R)
{
// Iterate over the range
for(int i = L; i <= R; i++)
{
int temp = i;
int c = 10;
int flag = 0;
// Iterate over the digits
while (temp > 0)
{
// Check if the current digit
// is >= the previous digit
if (temp % 10 >= c)
{
flag = 1;
break;
}
c = temp % 10;
temp /= 10;
}
// If the digits are in
// ascending order
if (flag == 0)
Console.Write(i + " ");
}
}
// Driver Code
public static void Main()
{
// Given range L and R
int L = 10, R = 15;
// Function call
printNum(L, R);
}
}
// This code is contributed by jrishabh99
java 描述语言
<script>
// Javascript program for the above approach
// Function to print all numbers
// in the range [L, R] having digits
// in strictly increasing order
function printNum(L, R)
{
// Iterate over the range
for(let i = L; i <= R; i++)
{
let temp = i;
let c = 10;
let flag = 0;
// Iterate over the digits
while (temp > 0)
{
// Check if the current digit
// is >= the previous digit
if (temp % 10 >= c)
{
flag = 1;
break;
}
c = temp % 10;
temp /= 10;
}
// If the digits are in
// ascending order
if (flag == 0)
document.write(i + " ");
}
}
// Driver code
// Given range L and R
let L = 10, R = 15;
// Function call
printNum(L, R);
// This code is contributed by sravan kumar
</script>
Output:
12 13 14 15
时间复杂度 O(N)，N 是 L 和 r 的绝对差 辅助空间: O(1)
版权属于：月萌API www.moonapi.com，转载请注明出处
