我正在尝试解决一个问题，而我的问题是，为什么我的解决方案不起作用？ 。这是问题，下面是答案。

问题来自leetcode:http://oj.leetcode.com/problems/decode-ways/

使用以下映射将包含A-Z字母的消息编码为数字:

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

public class Solution {
    public int numDecodings(String s) {
        if (s.isEmpty() || s.charAt(0) == '0') return 0;
        int decodings = 1;
        boolean used = false; // Signifies that the prev was already use as a decimal
        for (int index = s.length()-1 ; index > 0 ; index--) {
            char curr = s.charAt(index);
            char prev = s.charAt(index-1);
            if (curr == '0') {
                if (prev != '1' && prev != '2') {
                    return 0;
                }
                index--; // Skip prev because it is part of curr
                used = false;
            } else {
                if (prev == '1' || (prev == '2' && curr <= '6')) {
                    decodings = decodings * 2;
                    if (used) {
                        decodings = decodings - 1;
                    }
                    used = true;
                } else {
                    used = false;
                }
            }
        }
        return decodings;
    }
}

Input:"4757562545844617494555774581341211511296816786586787755257741178599337186486723247528324612117156948"
Output: 3274568
Expected: 589824

最佳答案

这是一个非常有趣的问题。首先，我将展示如何解决这个问题。我们将看到使用递归时并不会那么复杂，并且可以使用动态编程来解决该问题。我们将提供一种通用解决方案，该解决方案不对每个代码点的26上限进行硬编码。

关于术语的注释:我将使用术语代码点(CP)并不是Unicode的意思，而是通过1来指代其中一个代码编号26。每个代码点表示为可变数量的字符。我还将使用术语编码文本(ET)和明文(CT)的明显含义。当谈论序列或数组时，第一个元素称为head。其余元素是尾部。

理论前奏

static final int upperLimit  = 26;
static final int maxHeadSize = ("" + upperLimit).length();

static int numDecodings(String encodedText) {
    // check base case for the recursion
    if (encodedText.length() == 0) {
        return 1;
    }

    // sum all tails
    int sum = 0;
    for (int headSize = 1; headSize <= maxHeadSize && headSize <= encodedText.length(); headSize++) {
        String head = encodedText.substring(0, headSize);
        String tail = encodedText.substring(headSize);
        if (Integer.parseInt(head) > upperLimit) {
            break;
        }
        sum += numDecodings(tail);
    }

    return sum;
}

static final int upperLimit  = 26;
static final int maxHeadSize = ("" + upperLimit).length();

static int numDecodings(String encodedText) {
    return numDecodings(encodedText, new Integer[1 + encodedText.length()]);
}

static int numDecodings(String encodedText, Integer[] cache) {
    // check base case for the recursion
    if (encodedText.length() == 0) {
        return 1;
    }

    // check if this tail is already known in the cache
    if (cache[encodedText.length()] != null) {
        return cache[encodedText.length()];
    }

    // cache miss -- sum all tails
    int sum = 0;
    for (int headSize = 1; headSize <= maxHeadSize && headSize <= encodedText.length(); headSize++) {
        String head = encodedText.substring(0, headSize);
        String tail = encodedText.substring(headSize);
        if (Integer.parseInt(head) > upperLimit) {
            break;
        }
        sum += numDecodings(tail, cache);  // pass the cache through
    }

    // update the cache
    cache[encodedText.length()] = sum;
    return sum;
}

static final int upperLimit  = 26;
static final int maxHeadSize = ("" + upperLimit).length();

static int numDecodings(String encodedText) {
    int[] cache = new int[encodedText.length() + 1];

    // base case: the empty string at encodedText.length() is 1:
    cache[encodedText.length()] = 1;

    for (int position = encodedText.length() - 1; position >= 0; position--) {
        // sum directly into the cache
        for (int headSize = 1; headSize <= maxHeadSize && headSize + position <= encodedText.length(); headSize++) {
            String head = encodedText.substring(position, position + headSize);
            if (Integer.parseInt(head) > upperLimit) {
                break;
            }
            cache[position] += cache[position + headSize];
        }
    }

    return cache[0];
}

static int numDecodings(String encodedText) {
    // initialize the cache
    int[] cache = {1, 0, 0};

    for (int position = encodedText.length() - 1; position >= 0; position--) {
        // rotate the cache
        cache[2] = cache[1];
        cache[1] = cache[0];
        cache[0] = 0;

        // headSize == 1
        if (position + 0 < encodedText.length()) {
            char c = encodedText.charAt(position + 0);

            // 1 .. 9
            if ('1' <= c && c <= '9') {
                cache[0] += cache[1];
            }
        }

        // headSize == 2
        if (position + 1 < encodedText.length()) {
            char c1 = encodedText.charAt(position + 0);
            char c2 = encodedText.charAt(position + 1);

            // 10 .. 19
            if ('1' == c1) {
                cache[0] += cache[2];
            }
            // 20 .. 26
            else if ('2' == c1 && '0' <= c2 && c2 <= '6') {
                cache[0] += cache[2];
            }
        }
    }

    return cache[0];
}

关于java - 查看答案-解码方式，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/20342462/