跟隨者數字解碼
Problem statement:
問題陳述：
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.
給定一個僅包含I和D的模式 。 “ I”代表增加， “ D”代表減少。 設計一種算法以按照該模式打印最小數量。 數字為1-9 ，數字不能重復。
Example:
例：
Input:
IIDDIDD
Output:
12543876
Solution
解
The pattern & number to be generated
生成的圖案和編號
Length of minimum number = string length+1
最小數目 L的ength =字符串長度+ 1
Hence, maximum string length possible is 8, since we can construct only with different digits (1-9)
因此，最大字符串長度為8，因為我們只能用不同的數字(1-9)進行構造
'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”表示下一個數字比當前數字大1， “ D”表示下一個數字比當前數字小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'.
該問題可以在堆棧的幫助下使用。 概念是創建具有與本地連續序列“ D”的深度相同的連續數字的堆棧。
Prerequisite:
先決條件：
Input pattern, string s
輸入模式，字符串s
Stack st to store the digits
堆疊st以存儲數字
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 ++實現
#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
輸出量
First run:
enter pattern
IIDDIDD
The minimum number generated is:
12543876
Second run:
enter pattern
IIIIIIIIDDDDIII
Not possible,length>8
Example with explanation:
帶有說明的示例：
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
翻譯自: https://www.includehelp.com/icp/number-following-the-pattern.aspx
跟隨者數字解碼
