![](/img/trans.png)
[英]Dynamic Programming: Why does the code fail when I break the if statement up into 2 lines?
[英]Dynamic Programming - Word Break
我正在尝试解决这个问题。问题如下
给定一个输入字符串和一个单词词典,请找出输入字符串是否可以分割成以空格分隔的词典单词序列。
字典是一个字符串数组。
我的方法是在以下递归函数中存储递归调用的结果。 输出很好,但我发现从未使用过存储的结果。 我的解决方案通过了测试用例,希望是正确的,但是如果我知道是否使用DP,那将是很棒的。
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;
}
递归是强制性的吗? 我知道,迭代式DP解决方案最简单,最紧凑:
#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
您的代码实际上是错误的。 要使代码失败,请尝试输入“ likeman”
请注意,函数searchSentence
可能有两个不同的返回值,0或1。因此,如果将r
数组初始化为0,则无法保证r[x][y] = 0
时它是新状态。 用一些不可能的值(如-1或2)初始化r数组,然后再次测试。 现在,您可以轻松地确认,如果r[qbegin][qend] != -1
那么此状态已经过检查,因此您可以从此处返回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;
}
PS:有一些多余的代码行,但是我没有更改它们,而是在函数searchWord
的字符数组temp的末尾添加了一个空字符
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.