[LintCode/LeetCode] Decode Ways [String to Integer

andong777 發布于2019-08-14 15:44 / 2376人閱讀

摘要：用將子字符串轉化為，參見和的區別然后用動規方法表示字符串的前位到包含方法的個數。最后返回對應字符串末位的動規結果。

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將子字符串轉化為int，參見parseInt和valueOf的區別；
然后用動規方法：
dp[i]表示字符串s的前i位（0到i-1）包含decode方法的個數。
若前一位i-2到當前位i-1的兩位字符串s.substring(i-2, i)對應的數字lastTwo在10到26之間，則當前位dp[i]要加上這兩位字符之前一個字符對應的可能性：dp[i-2];
若當前位i-1的字符對應1到9之間的數字（不為0），則當前dp[i]要加上前一位也就是第i-2位的可能性：dp[i-1]。
最后返回對應字符串s末位的動規結果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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