题目
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” (1 2) or “L” (12).
The number of ways decoding “12” is 2.
分析
代码
/*---------------------------------------
* 日期:2015-06-23
* 作者:SJF0115
* 题目: 91.Decode Ways
* 网址:https://leetcode.com/problems/decode-ways/
* 结果:AC
* 来源:LeetCode
* 博客:
-----------------------------------------*/
#include <iostream>
using namespace std;
class Solution {
public:
int numDecodings(string s) {
int size = s.size();
if(s[0] == ‘0‘){
return 0;
}//if
if(size == 0 || size == 1){
return size;
}//if
int pre = 1,cur = 1,res;
for(int i = 1;i < size;++i){
if(isValid(s[i-1],s[i]) && isValid(s[i])){
res = pre + cur;
}//if
else if(!isValid(s[i-1],s[i]) && isValid(s[i])){
res = cur;
}//else
else if(isValid(s[i-1],s[i]) && !isValid(s[i])){
res = pre;
}//else
else{
return 0;
}//else
pre = cur;
cur = res;
}//for
return res;
}
private:
bool isValid(char pre,char cur){
if(pre == ‘1‘ || (pre == ‘2‘ && cur <= ‘6‘)){
return true;
}//if
return false;
}
bool isValid(char cur){
if(cur >= ‘1‘ && cur <= ‘9‘){
return true;
}//if
return false;
}
};
int main(){
Solution s;
string str("1202111110");
cout<<s.numDecodings(str)<<endl;
return 0;
}
运行时间
思路2–超时
/*---------------------------------------
* 日期:2015-06-21
* 作者:SJF0115
* 题目: 91.Decode Ways
* 网址:https://leetcode.com/problems/decode-ways/
* 结果:超时
* 来源:LeetCode
* 博客:
-----------------------------------------*/
#include <iostream>
using namespace std;
class Solution {
public:
int numDecodings(string s) {
int size = s.size();
if(size == 0 || size == 1){
return size;
}//if
int index = -1;
int count = 0;
helper(s,index,count,"");
return count;
}
private:
void helper(string &s,int index,int &count,string word){
int size = s.size();
if(index == size-1){
++count;
cout<<"word->"<<word<<endl;
return;
}//if
// 步长为1或者2
int num = 0;
for(int i = 1;i <= 2;++i){
if(index + i < size){
num = num * 10 + s[index+i] - ‘0‘;
if(num <= 26 && num > 0){
word += (‘A‘+num-1);
helper(s,index+i,count,word);
word.erase(word.size()-1);
}//if
else{
break;
}
}//if
}//for
}
};
int main(){
Solution s;
string str("1234");
cout<<s.numDecodings(str)<<endl;
return 0;
}
原文地址:http://blog.csdn.net/sunnyyoona/article/details/46611499