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上下文
这个问题是dailycodingproblem和leetcode问的

/* 给定映射 a = 1, b = 2, ... z = 26 和编码消息，计算可以解码的方式数。

例如，消息“111”将给出 3，因为它可以被解码为“aaa”、“ka”和“ak”。

您可以假设消息是可解码的。例如，不允许使用“001”。*/

var AlphabetCode = function(){};
AlphabetCode.prototype.decode = function(message) {
    //dp is optimal substructure, account for zero length string such as dp[0]
    let dp = Array(1+message.length).fill(0);
    //string is of length zero
    dp[0] = 0; // no ways to decode empty (tried toggling between 0 and 1)  
    //string is of length one
    dp[1] = parseInt(message[0]) == 0 ? 0 : 1; // there is no alphabet code for 0, 'a' starts at 1 

    //string is of length two or greater
    // go up to string length inclusive because index 0 is reserved for zero string
    for (let i = 2; i <= message.length; i++){
        let singleDigit = message.substring(i-1, i);
        let doubleDigit = message.substring(i-2, i);
        //console.log(singleDigit + ' ' + doubleDigit);
        //console.log(singleDigit[0]);
        if (1 <= parseInt(singleDigit) <= 9){
            dp[i] += dp[i-1];
            //console.log(dp[i]);
        }
        //console.log(doubleDigit[0]);
        if (doubleDigit[0] !='0' && 10 <= parseInt(doubleDigit) <= 26){
            //console.log('double valid');
            dp[i] += dp[i-2];
        }
    }
    // filled out the dp array and returning the accumulation of all subproblems
    return dp[message.length];
};
combinations = new AlphabetCode();
console.log('Number of ways to decode 10 are: (expect 1) ' + combinations.decode('10'));
console.log('Number of ways to decode 12 are: (expect 2) ' + combinations.decode('12'));
console.log('Number of ways to decode 226 are: (expect 3) ' + combinations.decode('226'));
console.log('Number of ways to decode 27 are: (expect 1) ' + combinations.decode('27'));

输出

Number of ways to decode 10 are: (expect 1) 1 
Number of ways to decode 12 are: (expect 2) 1
Number of ways to decode 226 are: (expect 3) 2
Number of ways to decode 27 are: (expect 1) 1

dp是最优子结构。尝试更改dp[0]为 0 或 1 以通过所有测试用例，但输出并不总是等于预期的数字。

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1 回答 1

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两个问题：

  • 正如您已经尝试过的那样，但dp[0]应该是 1，因为实际上空字符串是空消息的有效编码。所以很重要。

  • 你不能在 JavaScript 中做 python 风格的双重比较，所以这两个条件都是无效的：

    if (1 <= parseInt(singleDigit) <= 9)
if (doubleDigit[0] !='0' && 10 <= parseInt(doubleDigit) <= 26)

    将它们更改为；

    if (1 <= parseInt(singleDigit))
if (doubleDigit[0] !='0' && parseInt(doubleDigit) <= 26)

    删除的“条件”已经保证为真：当您已经验证第一个数字不是零时，一位数不能大于 9，而两位数至少为 10。

对代码的一些评论

创建一个构造函数是多余的，因为您不会在该“类”的实例中维护状态。而是创建一个以函数为成员的对象。

此外，您不需要dp为每个可能的字符串长度输入一个条目，因为您只需要“最后一个”两个结果来计算下一个，因此您可以只使用两个值。

最后，使用三元运算符和一元加号（用于转换为数字），您可以非常简洁地计算下一个计数：

const combinations = {
    decode(message) {
        // Only need memory of past two results (for string length - 1, and - 2)
        let dp = [1, 1];
        // No need to reject a starting zero, as the challenge says only valid input is given.
        for (let i = 2; i <= message.length; i++){
            let singleDigit = +message[i-1];
            let doubleDigit = +message.substring(i-2, i);
            dp = [
                dp[1], 
                (1 <= singleDigit) * dp[1]
                    + (10 <= doubleDigit && doubleDigit <= 26) * dp[0]
            ];
        }
        return dp[1];
    }
};

console.log('Number of ways to decode 10 are: (expect 1) ' + combinations.decode('10'));
console.log('Number of ways to decode 12 are: (expect 2) ' + combinations.decode('12'));
console.log('Number of ways to decode 226 are: (expect 3) ' + combinations.decode('226'));
console.log('Number of ways to decode 27 are: (expect 1) ' + combinations.decode('27'));

于 2019-07-19T20:13:49.483 回答