LeetCode-Decode Ways

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A message containing letters from A-Z is being encoded to numbers using the following mapping:

‘A‘ -> 1
‘B‘ -> 2
...
‘Z‘ -> 26

Given an encoded message containing digits, determine the total number of ways to decode it.

For example,
Given encoded message "12", it could be decoded as "AB" (1 2) or "L" (12).

The number of ways decoding "12" is 2.

    public int numDecodings(String s) {
        if (s == null || s.length() == 0 || s.charAt(0) == '0') return 0;
        int[] dp = new int[s.length()+1];    
        dp[0] = dp[1] = 1;
        for (int i = 1; i < s.length(); i++) {
            char pre = s.charAt(i-1);
            char c = s.charAt(i);
            if (c == '0') {
                if (pre > '2' || pre == '0') return 0;
                else dp[i+1] = dp[i-1];
            } else {
                dp[i+1] = dp[i];
                if (pre != '0' && (pre-'0')*10 + (c-'0') <= 26) dp[i+1] += dp[i-1];
            }
        }
        return dp[dp.length-1];
    }

此方法是从前往后递推，dp[i+1]是指以s.substring(0,i+1)字串的解，但是此代码比较繁杂，因为就是0这个坑。

看下面的代码：

    public int numDecodings(String s) {
        int n = s.length();
        if (n == 0) {
            return 0;
        }
        int[] table = new int[n+1];
        table[n] = 1;
        table[n-1] = s.charAt(n-1) != '0' ? 1 : 0;
        for (int i = n-2; i >= 0; i--) {
            if (s.charAt(i) == '0') {
                table[i] = 0;
            } else {
                int num = Integer.parseInt(s.substring(i, i+2));
                if (num <= 26) {
                    table[i] = table[i+1]+table[i+2];
                } else {
                    table[i] = table[i+1];
                }
            }
        }
        return table[0];
    }

另外一种思考问题的方法，从头到尾，还是从尾到头。一些问题逆向比较简单。

LeetCode-Decode Ways

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原文地址：http://blog.csdn.net/my_jobs/article/details/47981897