[leetcode]Decode Ways

动态规划。要注意way+=dp[i-1]或dp[i-2]以及way+=1的条件。我看有的人把数组命名为count的。

public class Solution {
    public int numDecodings(String s) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int len = s.length();
        if (len == 0) return 0;
        int[] dp = new int[len];

        for (int i = 0; i < len; i++) {
            int way = 0;
            if (s.charAt(i) != '0') {
                if (i-1 >= 0) {
                    way += dp[i-1];
                }
                else {
                    way += 1;
                }
            }
            if (i-1 >= 0 && (s.charAt(i-1) == '1' || (s.charAt(i-1) == '2' && s.charAt(i) <= '6'))) {
                if (i-2 >= 0) {
                    way += dp[i-2];
                }
                else {
                    way+= 1;
                }
            }
            dp[i] = way;
        }
        return dp[len-1];
    }
}

第二刷，做繁琐了

class Solution {
public:
    vector<int> ways;
    int numDecodings(string s) {
        if (s.size() == 0)
            return 0;
        ways.clear();
        ways.resize(s.size() + 1);
        for (int i = 0; i < ways.size(); i++)
        {
            ways[i] = -1;
        }
        return numDecodingsRe(s, 0);
    }
    
    int numDecodingsRe(const string &s, int start)
    {
        if (ways[start] != -1)
        {
            return ways[start];
        }
        if (start == s.size())
        {
            ways[start] = 1;
            return 1;
        }
        if (s[start] < '1' || s[start] > '9')
        {
            ways[start] = 0;
            return 0;
        }
        if (start == s.size() - 1)
        {
            ways[start] = 1;
            return 1;
        }
        if (s[start] == '1')
        {
            ways[start] = numDecodingsRe(s, start + 1) + numDecodingsRe(s, start + 2);
            return ways[start];
        }
        if (s[start] == '2' && s[start + 1] <= '6')
        {
            ways[start] = numDecodingsRe(s, start + 1) + numDecodingsRe(s, start + 2);
            return ways[start];
        }
        ways[start] = numDecodingsRe(s, start + 1);
        return ways[start];
    }
};

python3, dp, 注意单个的'0'没有对应编码

class Solution:
    def numDecodings(self, s: str) -> int:
        memo = {} # decode ways for s[ : i]
        for i in range(len(s) + 1): # i, length of sub str
            if i == 0:
                memo[i] = 1
            elif i == 1:
                if s[i - 1] == '0':
                    memo[i] = 0
                else:
                    memo[i] = 1
            else: # i >= 2
                nums = 0
                if int(s[i - 2:i]) <= 26 and int(s[i - 2:i]) >= 10:
                    nums += memo[i - 2]
                if s[i - 1] != '0':
                    nums += memo[i - 1]
                memo[i] = nums
        return memo[len(s)]