word-break

最新推荐文章于 2018-07-23 21:18:00 发布

njudongchen

最新推荐文章于 2018-07-23 21:18:00 发布

阅读量248

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode 文章标签： leetcode word-break

本文链接：https://blog.youkuaiyun.com/njudongchen/article/details/68954402

leetcode 专栏收录该内容

51 篇文章

订阅专栏

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.
For example, given
s =”leetcode”,
dict =[“leet”, “code”].
Return true because”leetcode”can be segmented as”leet code”.

这道题目可以利用动态规划的方法来做,首先设置dp[i]为以i结尾的是否能被字典分割. dp[i]和dp[j]和i-j之间的字符是否在字典中有关.

class Solution {
public:
    bool wordBreak(string s, unordered_set<string> &dict) {
    const unsigned int len = s.size();
    if (len == 0)
        return false;
    char* dp = new char[len];  //保存字符0-i之间是否能被字典分割.
    memset(dp, 0, sizeof(*dp)*len);
    int maxsize = 0;
    for (auto iter = dict.cbegin(); iter != dict.end(); ++iter)//计算字典中的最长字符串可以减少判断的次数
    {
        if (iter->size() > maxsize)
            maxsize = iter->size();
    }
    if (dict.find(s.substr(0, 1)) != dict.end())   //设置dp[0]
    {
        dp[0] = 1;
    }
    for (int i = 1; i != len; ++i)
    {
        if (i+1<= maxsize&&dict.find(s.substr(0, i + 1)) != dict.cend()) //判断是否从0-i在字典中,当然长度要小于字典中的最大长度.
        {   
            dp[i] = 1;
            continue;
        }
        for (int j = i - 1; j >= 0 && (i - j) <= maxsize; --j) //判断dp[j]为真和i-j在字典中是否同时成立
        {
            if (dp[j] == 1 && dict.find(s.substr(j+1, i - j)) != dict.cend()){
                dp[i] = 1;
                break;
            }
        }
    }
    bool res=dp[len-1];
    delete[] dp;  //释放空间
    return res;     
    }
};