跟随者数字解码_跟随模式的数字

跟随者数字解码

Problem statement:

问题陈述：

Given a pattern containing only I's and D's. 'I' stands for increasing and 'D' for decreasing. Devise an algorithm to print the minimum number following that pattern. Digits are from 1-9 and digits can't repeat.

给定一个仅包含I和D的模式 。 “ I”代表增加， “ D”代表减少。设计一种算法以按照该模式打印最小数量。数字为1-9 ，数字不能重复。

Example:

例：

    Input:
IIDDIDD  
Output:
12543876

Solution

解

The pattern & number to be generated

生成的图案和编号

Length of minimum number = string length+1
最小数目 L的ength =字符串长度+ 1

Hence, maximum string length possible is 8, since we can construct only with different digits (1-9)
因此，最大字符串长度为8，因为我们只能用不同的数字(1-9)进行构造
'I' means the next digit will be 1 greater than the current & 'D' means the next digit will be 1 less than the current digit
“ I”表示下一个数字比当前数字大1， “ D”表示下一个数字比当前数字小1。

"II" → 123
“ II”→123

"DD" → 321
“ DD”→321

The problem can be used with help of stack. The concept is to create stack with consecutive number same as depth of a local contiguous sequence of 'D'.

该问题可以在堆栈的帮助下使用。 概念是创建具有与本地连续序列“ D”的深度相同的连续数字的堆栈。

Prerequisite:

先决条件：

Input pattern, string s
输入模式，字符串s
Stack st to store the digits
堆叠st以存储数字

    Function findMinFromPattern(string s)
1.  Declare a stack st; 
2.  FOR i=0 : pattern length
EnQueue (st, i+1); //push i+1 at first, i+1 becuase i is 0-indexed 
IF (entire pattern is processed || s[i] =='I')
While(stack is not empty){  
Pop and print
End While
END IF
END FOR
END FUNCTION

C++ Implementation

C ++实现

#include <bits/stdc++.h>
using namespace std;
void findMinFromPattern(string s){
stack<int> st; //stack declared using STL
for(int i=0;i<=s.length();i++){
//push i+1 at first, i+1 becuase i is 0-indexed 
st.push(i+1); 
//when string length is processed or pattern in I
if(s.length()==i || s[i]=='I' ){ 
//pop and print until stack is empty
while(!st.empty()){ 
cout<<st.top();
st.pop();
}
}
}
cout<<endl;
}
int main(){
cout<<"enter pattern\n";    
string s;
cin>>s;
if(s.length()>8){
cout<<"Not possible,length>8\n";
}
else{
cout<<"The minimum number generated is:\n";
findMinFromPattern(s);//function to print
}
return 0;
}

Output

输出量

First run:
enter pattern
IIDDIDD
The minimum number generated is:
12543876
Second run:
enter pattern
IIIIIIIIDDDDIII
Not possible,length>8

Example with explanation:

带有说明的示例：

Pattern string:
IIDDIDD
0 th iteration
i=0
EnQueue(st,i+1)
Stack status:
1
S[i]=='I'
So pop and print until stack becomes empty
Thus printing:
1
Output up to now:
1
Stack status;
Empty stack
-------------------------------------------------------------
1st iteration
i=1
EnQueue(st,i+1)
Stack status:
2
S[i]=='I'
So pop and print until stack becomes empty
Thus printing:
2
Output up to now:
12
Stack status;
Empty stack
-------------------------------------------------------------
2nd iteration
i=2
EnQueue(st,i+1)
Stack status:
3
S[i]=='D'
Nothing to do
Output up to now:
12
Stack status;
3
-------------------------------------------------------------
3rd iteration
i=3
EnQueue(st,i+1)
Stack status:
3
4(top)
S[i]=='D'
Nothing to do
Output up to now:
12
Stack status;
3
4(top)
-------------------------------------------------------------
4th iteration
i=4
EnQueue(st,i+1)
Stack status:
3
4
5(top)
S[i]=='I'
So pop and print until stack becomes empty
Thus printing:
5, then 4,then 3
Output up to now:
12543
Stack status;
Empty stack
-------------------------------------------------------------
5th iteration
i=5
EnQueue(st,i+1)
Stack status:
6
S[i]=='D'
Nothing to do
Output up to now:
12543
Stack status;
6
-------------------------------------------------------------
6th iteration
i=6
EnQueue(st,i+1)
Stack status:
6
7(top)
S[i]=='D'
Nothing to do
Output up to now:
12543
-------------------------------------------------------------
7th iteration
i=7
EnQueue(st,i+1)
Stack status:
6
7
8(top)
Entire string is processed
Pop and print until stack becomes empty
Print:
8, then 7, then 6
Output up to now:
12543876
Exit:
Minimum no is:
12543876