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 lengthEnQueue (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 printEnd WhileEND IFEND FOREND FUNCTION

C++ Implementation

C ++實現

#include <bits/stdc++.h> using namespace std;void findMinFromPattern(string s){stack<int> st; //stack declared using STLfor(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 Iif(s.length()==i || s[i]=='I' ){ //pop and print until stack is emptywhile(!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: 12543876Second run: enter pattern IIIIIIIIDDDDIII Not possible,length>8 .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}} .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}}

Example with explanation:

帶有說明的示例：

Pattern string: IIDDIDD0 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

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