Word Break Problem 动态规划

最新推荐文章于 2025-01-19 02:09:00 发布

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本文介绍了一种解决单词拆分问题的算法实现。通过动态规划方法判断一个字符串是否能被拆分为字典中存在的单词序列。文章提供了完整的C++代码示例，并通过几个测试用例验证了算法的有效性。

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// Word Break Problem.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <iostream>
#include <string>

using namespace std;

int dictionaryContains(string word)
{
    string dictionary[] = {"mobile","samsung","sam","sung","man","mango",
                           "icecream","and","go","i","like","ice","cream"};
    int size = sizeof(dictionary)/sizeof(dictionary[0]);
    for (int i = 0; i < size; i++)
        if (dictionary[i].compare(word) == 0)
           return true;
    return false;
}

bool wordBreak(string str)
{
	int size = str.size();

	if (size == 0)  
		return true;
	
	bool *wb = new bool[size+1];

	for (int i = 0; i<= size; i++)
	{
		wb[i] = false;
	}
	wb[0] = true;

	for (int i = 0; i<size; i++)// idx of the inputed string
	{
		for (int j = 0; j<=i; j++)
		{
			if (wb[j] == true)
			{
				if (dictionaryContains(str.substr(j, i-j+1)))
				{
					wb[i+1] = true;
					break;
				}
			}
		}
	}

	bool retVal = wb[size] == true? true:false;

	delete [] wb;

	return retVal;
}


int _tmain(int argc, _TCHAR* argv[])
{
    wordBreak("ilikesamsung")? cout <<"Yes\n": cout << "No\n";
    wordBreak("iiiiiiii")? cout <<"Yes\n": cout << "No\n";
    wordBreak("")? cout <<"Yes\n": cout << "No\n";
    wordBreak("ilikelikeimangeiii")? cout <<"Yes\n": cout << "No\n";
    wordBreak("samsungandmango")? cout <<"Yes\n": cout << "No\n";
    wordBreak("samsungandmangok")? cout <<"Yes\n": cout << "No\n";
	return 0;
}