Leetcode 91. Decode Ways

A message containing letters from A-Z is being encoded to numbers using the following mapping:

‘A’ -> 1
‘B’ -> 2
…
‘Z’ -> 26
Given a non-empty string containing only digits, determine the total number of ways to decode it.

Example 1:

Input: “12”
Output: 2
Explanation: It could be decoded as “AB” (1 2) or “L” (12).
Example 2:

Input: “226”
Output: 3
Explanation: It could be decoded as “BZ” (2 26), “VF” (22 6), or “BBF” (2 2 6).

method dp

与Leetcode 70.爬楼梯相似，每一次状态的转移可能取决于先前的多个状态。
但比较坑的是测试用例里面有题目没说明的0情况

class Solution {
public:
    int numDecodings(string s) {
        if(s.size() == 0 || (s.size() == 1 && s[0] == '0')) return 0;
        if(s.size() == 1) return 1;
	    vector<int> dp(s.size() + 1, 0);
        dp[0] = 1;
        for(int i = 0; i < s.size(); ++i){
            dp[i+1] = s[i] == '0' ? 0 : dp[i];
            if(i > 0 && (s[i-1] == '1' || (s[i-1] == '2' && s[i] <= '6'))){
                dp[i+1] += dp[i-1];
            }
        }
        return dp.back();
    }
};

Similar questions:
62. Unique Paths
70. Climbing Stairs
509. Fibonacci Number

Practice them in a row for better understanding

summary

1. 每一次状态的转移可能取决于先前的多个状态。有的是在多个状态中取最优，有的是要在多个状态中取汇总。