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;
}
};