[LintCode/LeetCode] Decode Ways [String to Integer

andong777 发布于2019-08-14 15:44 / 2294人阅读

摘要：用将子字符串转化为，参见和的区别然后用动规方法表示字符串的前位到包含方法的个数。最后返回对应字符串末位的动规结果。

Problem

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.

Example

Given encoded message 12, it could be decoded as AB (1 2) or L (12).
The number of ways decoding 12 is 2.

Note

用parseInt将子字符串转化为int，参见parseInt和valueOf的区别；
然后用动规方法：
dp[i]表示字符串s的前i位（0到i-1）包含decode方法的个数。
若前一位i-2到当前位i-1的两位字符串s.substring(i-2, i)对应的数字lastTwo在10到26之间，则当前位dp[i]要加上这两位字符之前一个字符对应的可能性：dp[i-2];
若当前位i-1的字符对应1到9之间的数字（不为0），则当前dp[i]要加上前一位也就是第i-2位的可能性：dp[i-1]。
最后返回对应字符串s末位的动规结果dp[n]。

Solution updated 2018-9

class Solution {
    public int numDecodings(String s) {
        if (s == null || s.length() == 0) return 0;
        int n = s.length();
        int[] dp = new int[n+1];
        dp[0] = 1;                                                  //if the first two digits in [10, 26]
        dp[1] = s.charAt(0) == "0" ? 0 : 1;
        for (int i = 2; i <= n; i++) {
            if (s.charAt(i-1) != "0") dp[i] += dp[i-1];             //XXX5X
            int lastTwo = Integer.parseInt(s.substring(i-2, i));
            if (lastTwo >= 10 && lastTwo <= 26) dp[i] += dp[i-2];   //XXX10X
        }
        return dp[n];
    }
}

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