检查数字的位数是否交替增加和减少
给定一个数字N ，任务是检查数字的位数是否交替增加和减少。如果位数小于3 ，则返回false 。例如，如果 d1、d2 和 d3 是 N 的数字，则d1 < d2 > d3必须成立。
例子：
Input: N = 6834
Output: true
Explanation: Digits of N are increasing and decreasing alternatively, i.e 6 < 8 > 3 < 4
Input: N = 32
Output: false
方法：这个任务可以通过将给定的整数转换成字符串来解决。遍历字符串并检查相邻字符是否符合给定条件。
下面是上述方法的实现：
C++
// C++ implementation for the above approach
#include
using namespace std;
// Utility function to check if
// the digits of the current
// integer forms a wave pattern
bool check(int N)
{
// Convert the number to a string
string S = to_string(N);
// Loop to iterate over digits
for (int i = 0; i < S.size(); i++) {
if (i == 0) {
// Next character of
// the number
int next = i + 1;
// Current character is
// not a local minimum
if (next < S.size()) {
if (S[i] >= S[next]) {
return false;
}
}
}
else if (i == S.size() - 1) {
// Previous character of
// the number
int prev = i - 1;
if (prev >= 0) {
// Character is a
// local maximum
if (i & 1) {
// Character is not
// a local maximum
if (S[i] <= S[prev]) {
return false;
}
}
else {
// Character is a
// local minimum
if (S[i] >= S[prev]) {
return false;
}
}
}
}
else {
int prev = i - 1;
int next = i + 1;
if (i & 1) {
// Character is a
// local maximum
if ((S[i] > S[prev])
&& (S[i] > S[next])) {
}
else {
return false;
}
}
else {
// Character is a
// local minimum
if ((S[i] < S[prev])
&& (S[i] < S[next])) {
}
else {
return false;
}
}
}
}
return true;
}
// Driver Code
int main()
{
int N = 64;
cout << (check(N) ? "true" : "false");
} Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
// Utility function to check if
// the digits of the current
// integer forms a wave pattern
static Boolean check(int N)
{
// Convert the number to a string
String S = Integer.toString(N);
// Loop to iterate over digits
for (int i = 0; i < S.length(); i++) {
if (i == 0) {
// Next character of
// the number
int next = i + 1;
// Current character is
// not a local minimum
if (next < S.length()) {
if (S.charAt(i) >= S.charAt(next)) {
return false;
}
}
}
else if (i == S.length() - 1) {
// Previous character of
// the number
int prev = i - 1;
if (prev >= 0) {
// Character is a
// local maximum
if (i % 2 == 1) {
// Character is not
// a local maximum
if (S.charAt(i) <= S.charAt(prev)) {
return false;
}
}
else {
// Character is a
// local minimum
if (S.charAt(i) >= S.charAt(prev)) {
return false;
}
}
}
}
else {
int prev = i - 1;
int next = i + 1;
if (i % 2 == 1) {
// Character is a
// local maximum
if ((S.charAt(i) > S.charAt(prev))
&& (S.charAt(i) > S.charAt(next))) {
}
else {
return false;
}
}
else
{
// Character is a
// local minimum
if ((S.charAt(i) < S.charAt(prev))
&& (S.charAt(i) < S.charAt(next))) {
}
else {
return false;
}
}
}
}
return true;
}
// Driver Code
public static void main (String[] args) {
int N = 64;
if(check(N) == true)
System.out.print("true");
else
System.out.print("false");
}
}
// This code is contributed by hrithikgarg03188Python3
# Python implementation for the above approach
# Utility function to check if
# the digits of the current
# integer forms a wave pattern
def check(N):
# Convert the number to a string
S = str(N)
# Loop to iterate over digits
for i in range(len(S)):
if (i == 0):
# Next character of
# the number
next = i + 1
# Current character is
# not a local minimum
if (next < len(S)):
if (S[i] >= S[next]):
return False
elif (i == len(S) - 1):
# Previous character of
# the number
prev = i - 1
if (prev >= 0):
# Character is a
# local maximum
if (i & 1):
# Character is not
# a local maximum
if (S[i] <= S[prev]):
return False
else:
# Character is a
# local minimum
if (S[i] >= S[prev]):
return False
else:
prev = i - 1
next = i + 1
if (i & 1):
# Character is a
# local maximum
if ((S[i] > S[prev]) and (S[i] > S[next])):
print("", end="")
else:
return False
else:
# Character is a
# local minimum
if ((S[i] < S[prev]) and (S[i] < S[next])):
print("", end="")
else:
return False
return True
# Driver Code
N = 64
print("true" if check(N) else "false")
# This code is contributed by Saurabh JaiswalC#
// C# program for the above approach
using System;
class GFG {
// Utility function to check if
// the digits of the current
// integer forms a wave pattern
static bool check(int N)
{
// Convert the number to a string
string S = N.ToString();
// Loop to iterate over digits
for (int i = 0; i < S.Length; i++) {
if (i == 0) {
// Next character of
// the number
int next = i + 1;
// Current character is
// not a local minimum
if (next < S.Length) {
if (S[i] >= S[next]) {
return false;
}
}
}
else if (i == S.Length - 1) {
// Previous character of
// the number
int prev = i - 1;
if (prev >= 0) {
// Character is a
// local maximum
if (i % 2 == 1) {
// Character is not
// a local maximum
if (S[i] <= S[prev]) {
return false;
}
}
else {
// Character is a
// local minimum
if (S[i] >= S[prev]) {
return false;
}
}
}
}
else {
int prev = i - 1;
int next = i + 1;
if (i % 2 == 1) {
// Character is a
// local maximum
if ((S[i] > S[prev])
&& (S[i] > S[next])) {
}
else {
return false;
}
}
else
{
// Character is a
// local minimum
if ((S[i] < S[prev])
&& (S[i] < S[next])) {
}
else {
return false;
}
}
}
}
return true;
}
// Driver Code
public static void Main () {
int N = 64;
if(check(N) == true)
Console.Write("true");
else
Console.Write("false");
}
}
// This code is contributed by Samim Hossain Mondal.Javascript
输出
false时间复杂度：O(N)，N是位数
辅助空间：O(N)