91. Decode Ways--leetcode

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/u014491519/article/details/78488330

题目

‘A’ -> 1
‘B’ -> 2
‘Z’ -> 26
Given encoded message “12”, it could be decoded as “AB” (1 2) >or “L” (12).
The number of ways decoding “12” is 2.

思路：动态规划
dp[i]表示前i个数字的解码方式个数

第一次的思路

public int numDecodings(String s) {
        if(s.length() == 0)
                return 0;
            if(s.charAt(0) == '0')
                return 0;
            int[] dp = new int[s.length()];
            dp[0] = 1;
            for(int i = 1; i < s.length();i++){
                if(s.charAt(i) == '0'){
                    if(s.charAt(i-1) == '1' || s.charAt(i-1) == '2')
                        dp[i] = dp[i-1];
                    else
                        return 0;
                }else if(s.charAt(i) == '1' || s.charAt(i) == '2'){
                    if(i < s.length()-1){
                        if(s.charAt(i+1) == '0')
                            dp[i] = dp[i-1];
                        else{
                            if(s.charAt(i-1) == '1' || s.charAt(i-1) == '2')
                                dp[i] = dp[i-1] + 1;
                            else
                                dp[i] = dp[i-1];
                        }
                    }else{
                        if(s.charAt(i-1) == '1' || s.charAt(i-1) == '2')
                            dp[i] = dp[i-1] + 1;
                        else
                            dp[i] = dp[i-1];
                    }
                }else if(s.charAt(i) <= '6'){
                    if(s.charAt(i-1) == '1' || s.charAt(i-1) == '2')
                        dp[i] = dp[i-1] + 1;
                    else
                        dp[i] = dp[i-1];
                }else{
                    if(s.charAt(i-1) == '1')
                        dp[i] = dp[i-1] + 1;
                    else
                        dp[i] = dp[i-1];
                }
            }
            return dp[s.length()-1];
    }

上述代码有错误，计算的方式不对

正确的代码：

 public int numDecodings(String s) {
        if(s.length() == 0)
            return 0;
        if(s.charAt(0) == '0')
            return 0;
        int[] dp = new int[s.length()];
        dp[0] = 1;
        for(int i = 1; i < s.length();i++){
           if(s.charAt(i) == '0'){
                if(s.charAt(i-1) == '0' || s.charAt(i-1) >= '3')
                    return 0;
                else
                    dp[i] = (i>1)?dp[i-2]:1;
            }else{
                if(s.charAt(i-1) == '0' || s.charAt(i-1) > '2') dp[i] = dp[i-1];
                else if(s.charAt(i-1) == '2' && s.charAt(i) > '6') dp[i] = dp[i-1];
                else {
                    dp[i] = (i>1)?dp[i-2]:0;
                    dp[i] += (i>1)?dp[i-1]:(dp[i-1]+1);
                }
            }
        }
        return dp[s.length()-1];
    }

思路：待补……