给定字符串连续数字和一个数字 Y，任务是找到最小集合的数量，使得每个集合都遵循以下规则：
- 集合应包含连续数字
- 任何数字都不能多次使用。
- 集合中的数量不应超过 Y。
例子：
Input: s = "1234", Y = 30
Output: 3
Three sets of {12, 3, 4}
Input: s = "1234", Y = 4
Output: 4
Four sets of {1}, {2}, {3}, {4} 方法：可以使用贪心方法解决以下问题，其步骤如下：
- 在字符串迭代，并使用 num*10 + (s[i]-‘0’) 将数字连接到一个变量（比如 num）
- 如果数字不大于 Y，则标记 f = 1。
- 如果数量超过 Y，则在 f = 1 时增加计数并将 f 重新初始化为 0，并将 num 初始化为 s[i]-‘0’ 或 num 为 0。
- 在字符串完全迭代后，如果 f 为 1，则增加计数。
下面是上述方法的实现：
C++
// C++ program to find the minimum number
// sets with consecutive numbers and less than Y
#include
using namespace std;
// Function to find the minimum number of shets
int minimumSets(string s, int y)
{
// Variable to count
// the number of sets
int cnt = 0;
int num = 0;
int l = s.length();
int f = 0;
// Iterate in the string
for (int i = 0; i < l; i++) {
// Add the number to string
num = num * 10 + (s[i] - '0');
// Mark that we got a number
if (num <= y)
f = 1;
else // Every time it exceeds
{
// Check if previous was
// anytime less than Y
if (f)
cnt += 1;
// Current number
num = s[i] - '0';
f = 0;
// Check for current number
if (num <= y)
f = 1;
else
num = 0;
}
}
// Check for last added number
if (f)
cnt += 1;
return cnt;
}
// Driver Code
int main()
{
string s = "1234";
int y = 30;
cout << minimumSets(s, y);
return 0;
} Java
// Java program to find the minimum number
// sets with consecutive numbers and less than Y
import java.util.*;
class solution
{
// Function to find the minimum number of shets
static int minimumSets(String s, int y)
{
// Variable to count
// the number of sets
int cnt = 0;
int num = 0;
int l = s.length();
boolean f = false;
// Iterate in the string
for (int i = 0; i < l; i++) {
// Add the number to string
num = num * 10 + (s.charAt(i) - '0');
// Mark that we got a number
if (num <= y)
f = true;
else // Every time it exceeds
{
// Check if previous was
// anytime less than Y
if (f)
cnt += 1;
// Current number
num = s.charAt(i) - '0';
f = false;
// Check for current number
if (num <= y)
f = true;
else
num = 0;
}
}
// Check for last added number
if (f == true)
cnt += 1;
return cnt;
}
// Driver Code
public static void main(String args[])
{
String s = "1234";
int y = 30;
System.out.println(minimumSets(s, y));
}
}
// This code is contributed by
// Shashank_SharmaPython3
# Python3 program to find the minimum number
# sets with consecutive numbers and less than Y
import math as mt
# Function to find the minimum number of shets
def minimumSets(s, y):
# Variable to count the number of sets
cnt = 0
num = 0
l = len(s)
f = 0
# Iterate in the string
for i in range(l):
# Add the number to string
num = num * 10 + (ord(s[i]) - ord('0'))
# Mark that we got a number
if (num <= y):
f = 1
else: # Every time it exceeds
# Check if previous was anytime
# less than Y
if (f):
cnt += 1
# Current number
num = ord(s[i]) - ord('0')
f = 0
# Check for current number
if (num <= y):
f = 1
else:
num = 0
# Check for last added number
if (f):
cnt += 1
return cnt
# Driver Code
s = "1234"
y = 30
print(minimumSets(s, y))
# This code is contributed by
# Mohit kumar 29C#
// C# program to find the minimum number
// sets with consecutive numbers and less than Y
using System;
class solution
{
// Function to find the minimum number of shets
static int minimumSets(string s, int y)
{
// Variable to count
// the number of sets
int cnt = 0;
int num = 0;
int l = s.Length ;
bool f = false;
// Iterate in the string
for (int i = 0; i < l; i++) {
// Add the number to string
num = num * 10 + (s[i] - '0');
// Mark that we got a number
if (num <= y)
f = true;
else // Every time it exceeds
{
// Check if previous was
// anytime less than Y
if (f)
cnt += 1;
// Current number
num = s[i] - '0';
f = false;
// Check for current number
if (num <= y)
f = true;
else
num = 0;
}
}
// Check for last added number
if (f == true)
cnt += 1;
return cnt;
}
// Driver Code
public static void Main()
{
string s = "1234";
int y = 30;
Console.WriteLine(minimumSets(s, y));
}
// This code is contributed by Ryuga
}PHP
Javascript
输出：
3