[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];

    }

};