LeetCode 91. Decode Ways

最新推荐文章于 2020-02-27 10:19:39 发布

原创最新推荐文章于 2020-02-27 10:19:39 发布 · 112 阅读

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本文介绍了一种使用动态规划解决数字字符串解码问题的方法。给定一个仅包含数字的非空字符串，确定解码该字符串的所有可能方式的总数。例如，字符串12可以解码为AB或L，因此解码方式有两种。文章详细解释了动态规划的实现过程。

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

'A' -> 1
'B' -> 2
...
'Z' -> 26

Given a non-empty string containing only digits, determine the total number of ways to decode it.

Example 1:

Input: "12"
Output: 2
Explanation: It could be decoded as "AB" (1 2) or "L" (12).

Example 2:

Input: "226"
Output: 3
Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).

用动态规划解决：

    public int numDecodings(String s) {
        int[] dp = new int[s.length()+1];//dp[i]表示s[0,i-1]有多少种划分方法
        dp[0] = 1;//为了dp[2]
        dp[1] = s.charAt(0)!='0'?1:0;
                                
        for(int i = 2;i<dp.length;i++){//dp[i]表示s[0,i-1]
            char one = s.charAt(i-1);
            if(one!='0'){
              dp[i] +=  dp[i-1];
            }
            int two = Integer.valueOf(s.substring(i-2,i));//s[i-2,i-1]
            if(two>=10 && two<=26){
                dp[i] += dp[i-2];
            }
        }
        return dp[dp.length-1];
    }