Leetcode 91. Decode Ways

Leetcode 91. Decode Ways

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题目描述

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.

输入：数组字符串s
输出：可能的字母组成个数

思路

动态规划：用f[i]表示到第i个字符时的总共可能组成个数，类似于爬楼梯，当前的位置可能是前1格过来的，也可能是前2格过来的。另外，需要一些判断语句处理临界条件。
状态转移方程：

$$f[i]=\begin{cases} f[i-2]& \text{ if } s[i]=0,0<s[i-1]<3 \\ f[i-1]& \text{ if } s[i]\neq 0, 26<s[i-1,i] \\ f[i-1] + f[i-2]& \text{ if } s[i]\neq 0, 0<s[i-1,i]\leq26 \\ \end{cases}$$
那么f[n]就是答案，复杂度 $O(n)$

代码

class Solution {
public:
    int numDecodings(string s) {
        if(s.length()==0) return 0;
        if(s[0]=='0') return 0;
        int f[s.length()] = {0};
        f[0]=1;
        for(int i = 1;i<s.length();i++){
            if(s[i]=='0'){
                if((s[i-1]=='1'||s[i-1]=='2')){
                    if(i>=2)
                        f[i] = f[i-2];
                    else
                        f[i] = 1;
                }else
                    return 0;
            }else{
                if(s.substr(i-1,2)<="26"&&s[i-1]!='0'){
                    if(i>=2)
                        f[i] = f[i-1] + f[i-2];
                    else
                        f[i] = f[i-1] + 1;
                }else
                    f[i] = f[i-1];
            }
        }

        return f[s.length()-1];
    }
};