LeetCode 91 动态规划 Decode Ways 解码方法

Posted 2020-12-10 Muche

LeetCode 91 动态规划 Decode Ways 解码方法

LeetCode

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).

dp[i] 以 s[i] 结尾的前缀子串有多少种解码方法
dp[i] = dp[i - 1] * 1 if s[i] != ‘0‘
dp[i] += dp[i - 2] * 1 if 10 <= int(s[i - 1..i]) <= 26
这里的关键还是由哪个元素过来的，然后定义好dp(i)的定义
参考

class Solution {
    public int numDecodings(String s) {
        int n = s.length();
        if(n == 0) return 0;
        if(s.charAt(0) == ‘0‘) return 0;
        int[] dp = new int[n];
        dp[0] = 1;
        for(int i = 1; i < n; i++){
            int first = s.charAt(i-1)-‘0‘;
            int second = s.charAt(i)-‘0‘;       
            //从i-1过来，相等
            if(second != 0){
                dp[i] = dp[i-1];
            }
            
            //从i-2过来，保证两个数组合起来为合理的解码即可。这个时候相当于有两个分支，所以是加
            int tmp = first*10+second;
            if(tmp >= 10 && tmp <= 26){
                if(i == 1)
                    dp[i] += 1;
                else
                    dp[i] += dp[i-2];
            }
        }
        return dp[n-1];
    }
}