LeetCode 91. Decode Ways

1. 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.

2. 解题思路

这实际上就是一个动态规划问题, 记 S[i] 为到第i个位置时候， 有多少种解析数量。 则

S[i]=(第一位是否为0:S[i−1]:0)+(后两位是否可以构成1∼26？S[i−2]：0);

3. code

class Solution {
public:
    int numDecodings(string s) {
        if (s.size() == 0)
            return 0;

        vector<int> arr(s.size() + 1, 0);
        arr[0] = 1;
        for (int i = 1; i < (int)arr.size(); i++){
            if (s[i - 1] != '0')
                arr[i] = arr[i - 1];
            if (i > 1 && (s[i - 2] == '1' || s[i - 2] == '2' && s[i - 1] <= '6'))
                arr[i] += arr[i - 2];
        }
        return arr[s.size()];
    }
};

4. 大神解法

int numDecodings(string s) {
    if (!s.size() || s.front() == '0') return 0;
    // r2: decode ways of s[i-2] , r1: decode ways of s[i-1] 
    int r1 = 1, r2 = 1;

    for (int i = 1; i < s.size(); i++) {
        // zero voids ways of the last because zero cannot be used separately
        if (s[i] == '0') r1 = 0;

        // possible two-digit letter, so new r1 is sum of both while new r2 is the old r1
        if (s[i - 1] == '1' || s[i - 1] == '2' && s[i] <= '6') {
            r1 = r2 + r1;
            r2 = r1 - r2;
        }

        // one-digit letter, no new way added
        else {
            r2 = r1;
        }
    }

    return r1;
}