Word Break

Word Break

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. You may assume the dictionary does not contain duplicate words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

UPDATE (2017/1/4):
The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

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解題技巧：

對於該問題，採用動態規劃的方法求解，用數組isBreak表示第i個位置能否分割，那麼狀態轉移方程有：

isBreak[i] = isBreak[j] && s. substr(j, i-j)在單詞表中存在 其中，0<= j < i

代碼：

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;

//利用動態規劃求解
bool wordBreak(string s, vector<string>& wordDict)
{
    int len = s.size();
    sort(wordDict.begin(), wordDict.end());
    vector<bool> isBreak(len+1, false);
    isBreak[0] = true;
    int flag = 0;
    for(int i = 1; i <= len; i ++)
    {
        for(int j = i - 1; j >= 0; j --)
        {
            flag = 0;
            if(isBreak[j])
            {
                string tmp = s.substr(j, i-j);
                vector<string>::iterator it = wordDict.begin();
                while(it != wordDict.end())
                {
                    if(*it == tmp)
                    {
                        isBreak[i] = true;
                        flag = 1;
                        break;
                    }
                    else if(*it > tmp) break;

                    it ++;
                }
            }
            if(flag == 1) break;
        }
    }
    return isBreak[len];
}

int main()
{
    string s, word;
    vector<string> wordDict;
    cin >> s;
    while(cin >> word)
        wordDict.push_back(word);
    cout<<wordBreak(s, wordDict);
}