LeetCode 91. Decode Ways

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There are two ways to solve this problem. One is to solve it front-back, the other is back-front.

I currently only wrote the back-front method:

Suppose we have such a string:    a0, a1, a2.... ai, ai+1, ai+2...an-1

Suppose we have get that from an-1, an-2...ai+2, the ways of decoding such a sequence is dp[i+2]

There are three conditions to consider:

1: if ai == 1, we dont need to care about ai+1, the ways to decode is 1+ dp[i+2];

2: if ai == 2, ai+1 should be smaller than ‘6‘, the ways to decode is 1 + dp[i+2].

3: if ai == 2, ai+1 greater than ‘6‘, the ways to decode is dp[i+1].

Think that, decode a individual number is 1, but 0 has no way to decode, thus, 0.

So, we can have the dynamic function:  dp[i] = dp[i] + dp[i+2] if(s[i] == 1 || s[i] == 2 && s[i + 1] < ‘6‘)

#include <string>
#include <vector>
#include <iostream>
using namespace std;

/*
  A message containing letters from A-Z is being encoded to numbers using the following mapping:
  'A' --> 1
  'B' --> 2
  ....
  'Z' --> 26
  Given an encoded message containing digits, determine the total number of ways to decode it.
  For example:
  Given encoded message "12", it could be decoded as "AB" or "L". 
  The number of ways decoding "12" is 2.
*/
/*
  Analyze: 
  Several example will make this question clear.
  "102" --> "10", "2" -> total method is 1
  "12" --> {"1", "2"}, "12" --> total method is 2.
  "127" --> {"1", "2", "7"}, {"12", "7"} --> total method is 2
  "126" --> {"1", "2", "6"}, {"12", "6"}, {"1", "26"}--> total method is 3.
  There are several cases need special attentions.
  1: "1"/"2" + "0", the char "0" can be interpreted as a decode way.
  2: "2" + "7, 8, 9" can only be interpreted as one way.
*/

int numDecodings(string s) {
  int n = s.size();
  vector<int> dp(n+2, 1);
  for(int i = s.size() - 1; i >= 0; --i) {
    if(s[i] == 0) dp[i] = 0;
    else dp[i] = dp[i + 1];

    if(i + 1 < s.size() && (s[i] == '1' || (s[i] == '2' && s[i+1] <= '6'))) {
      dp[i] += dp[i+2];
    }
  }
  return dp[0];
}



int main(void) {
  cout << "123" << " decode ways: "<< numDecodings("123") << endl;
  cout << "102" << " decode ways: "<< numDecodings("102") << endl;
  cout << "127" << " decode ways: "<< numDecodings("127") << endl;
}

LeetCode 91. Decode Ways

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原文地址：http://blog.csdn.net/github_34333284/article/details/51352223