Question

我正在努力解决this问题。问题如下给定输入字符串和单词字典，找出输入字符串是否可以分割成空格分隔的字典单词序列。

Dictionary是一个字符串数组。

我的方法是以下递归fn，存储递归调用的结果。输出很好，但我发现存储的结果从未使用过。我的解决方案在通过测试用例时是有道理的。但如果我知道是否使用了DP，我会很高兴。

代码是：

#include <iostream>
#include <string.h>
using namespace std;

int r[100][100] = {0};  //To Store the calculated values


bool searchWord(char q[], char D[][20], int start, int end) {
    cout << "In Search Word Loop with " << start << " " << end << endl;
    char temp[end - start + 1];
    int j = 0;

    for (int i = start; i <= end ; ++i) {
        //cout << "Looping i " << i << endl;
        temp[j] = q[i];
        j++;
    }

    // cout << "For Word " << temp << endl;
    for (int i = 0; i < 12; ++i) {
        // cout << "Comparing with " << D[i] << endl;
        if (!strcmp(temp, D[i])) {
            cout << "Found Word" << temp << " " << D[i] << endl;
            return 1;
        }
    }

    return 0;
}

bool searchSentence(char q[], char D[][20], int qstart, int qend) {
    cout << "In Search Sentence Loop" << endl;
    if (r[qstart][qend] != 0) {
        cout << "DP Helped!!!" << endl;
        return 1;
    }

    if (qstart == qend) {
        if (searchWord(q, D, qstart, qstart))
            return 1;
        else return 0;
    }
    if (qstart > qend) return 1;

    int i;
    for (i = qstart; i <= qend; i++) {
        if (searchWord(q, D, qstart, i)) {
            r[i + 1][qend] = searchSentence(q, D, i + 1, qend);
            if (r[i + 1][qend] == 1) return 1;
        }
    }

    return 0;
}

int main() {
    char D[20][20] = { "i", "like", "sam", "sung", "samsung", "mobile", "ice", "cream", "icecream", "man", "go", "mango"};
    char q[100] = "samsungmango";

    int index = 0; char ch;
    ch = q[0];
    while (ch != '\0') {
        index++;
        ch = q[index];
    }

    if (searchSentence(q, D, 0, index - 1))
        cout << "Yes" << endl;
    else cout << "No" << endl;
}

Answer 1

递归是强制性的吗？我知道，迭代DP解决方案最简单紧凑：

#include <stdio.h>
#include <string.h>

int main() {
  const char *D[] = { "i", "like", "sam", "sung", "samsung", "mobile", "ice", "cream", "icecream", "man", "go", "mango", NULL};
  const char q[] = "samsungmango";
  char dp[100];
  short d_len[20];
  memset(dp, 0, sizeof(dp));
  dp[0] = 1; // 0 element is always reacheable
  int i, j;
  // compute dict string lengths
  for(i = 0; D[i]; i++)
      d_len[i] = strlen(D[i]);

  // Compute splits using DP array
  for(i = 0; q[i] != 0; i++)
      if(dp[i])  // this index is reacheable
          for(j = 0; D[j]; j++) // try to make next reacheable indexes
              if(strncmp(&q[i], D[j], d_len[j]) == 0)
                  dp[i + d_len[j]] = 1; // That position is reacheable, too

  // if EOLN(q) is reached, then yes
  printf("Answer is %s\n", dp[i]? "YES" : "NO");

} // main

Answer 2

您的代码实际上是错误的。要使代码失败，请尝试像＆＃34; likeman＆＃34;

这样的输入

请注意，函数searchSentence，0或1有两种不同的返回值。因此，如果您使用0初始化r数组，则无法保证它是新的r[x][y] = 0时的状态。使用一些不可能的值（例如-1或2）为该程序初始化r数组并再次测试。现在您可以轻松确认，如果r[qbegin][qend] != -1已经检查了此状态，那么您可以从此处返回r[qbegin][qend]

更新的代码：

#include <iostream>
#include <string.h>
using namespace std;

int r[100][100];  //To Store the calculated values

bool searchWord(char q[], char D[][20], int start, int end)
{
    cout << "In Search Word Loop with " << start << " " << end << endl;



    char temp[end - start + 1];
    int j = 0;

    for (int i = start; i <= end ; ++i)
    {
        //cout << "Looping i " << i << endl;
        temp[j] = q[i];
        j++;
    }
    temp[j] = '\0';

    //cout << "For Word " << temp << endl;

    for (int i = 0; i < 12; ++i)
    {
        // cout << "Comparing with " << D[i] << endl;
        if (!strcmp(temp, D[i]))
        {
            cout << "Found Word" << temp << " " << D[i] << endl;
            return 1;
        }
    }

    return 0;
}

bool searchSentence(char q[], char D[][20], int qstart, int qend)
{
    cout << "In Search Sentence Loop" << endl;

    if (r[qstart][qend] != -1)
    {
        cout << "DP Helped!!!" << endl;
        return r[qstart][qend];
    }


    if (qstart == qend)
    {
        if (searchWord(q, D, qstart, qstart))
            return 1;
        else return 0;
    }
    if (qstart > qend) return 1;

    int i;

    for (i = qstart; i <= qend; i++)
    {
        if (searchWord(q, D, qstart, i))
        {
            r[i + 1][qend] = searchSentence(q, D, i + 1, qend);
            if (r[i + 1][qend] == 1) return 1;
        }
    }
    return 0;
}

int main()
{
    char D[20][20] = { "i", "like", "sam", "sung", "samsung", "mobile", "ice", "cream", "icecream", "man", "go", "mango"};
    char q[100] = "ilike";

    int index = 0; char ch;
    ch = q[0];
    memset(r, -1, sizeof(r));
    while (ch != '\0')
    {
        index++;
        ch = q[index];
    }

    if (searchSentence(q, D, 0, index - 1))
    cout << "Yes" << endl;
    else cout << "No" << endl;
}

P.S：有一些冗余的代码行，但我没有改变它们，我在函数searchWord中的字符数组temp的末尾添加了一个空字符

动态编程 - 断言

2 个答案: