lintcode 解码方法

简单的动态规划

class Solution {
public:
    
    /*
     * @param s: a string,  encoded message
     * @return: an integer, the number of ways decoding
     */
    int numDecodings(string s) {
        // write your code here
        if(s == "" || s[0] == '0') return 0;
        int dp[110];

        memset(dp, 0, sizeof(dp));
        dp[0] = 1;
        for(int i = 1; i < s.length(); ++i){
                if(s[i] == '0'){
                    if(s[i - 1] != '1' && s[i -1] != '2') return 0;
                    else {
                        dp[i] = dp[i] + (i - 2 >= 0 ? dp[i-2] : 1);
                    }
                } else {
                    if(s[i - 1] == '1'){
                        dp[i] = dp[i] + (i - 2 >= 0 ? dp[i-2] : 1);
                        dp[i] = dp[i] + dp[i - 1];
                    } else if(s[i - 1] == '2'){
                        if(s[i] <= '6')
                            dp[i] = dp[i] + (i - 2 >= 0 ? dp[i-2] : 1);
                        dp[i] = dp[i] + dp[i - 1];
                    } else {
                        dp[i] = dp[i] + dp[i - 1];
                    }
                }         
        }

        return dp[s.length() - 1];
    }
};