单词切分

给出一个字符串s和一个词典，判断字符串s是否可以被空格切分成一个或多个出现在字典中的单词。

样例

给出
s = "lintcode"
dict = ["lint","code"]
返回 true 因为"lintcode"可以被空格切分成"lint code"

1.我的答案 递归。。。所以超时了。。。简直不懂动态规划的套路。。哭/(ㄒoㄒ)/~~

class Solution {
public:
/**
* @param s: A string s
* @param dict: A dictionary of words dict
*/
bool partitions(string s, unordered_set<string> &dict, int num){
if(dict.find(s) != dict.end())
return true;
if(s.size() == num)
return false;
for(int i = 0; i < s.size(); i++){
string s1 = s.substr(0, i+1);
string s2 = s.substr(i+1, s.size() - i);
bool b1 = partitions(s1, dict, i+1);
if(b1 == false)
continue;
bool b2 = partitions(s2, dict, s.size() - i);
if(b1 && b2)
return true;
}
return false;
}

bool wordBreak(string s, unordered_set<string> &dict) {
// write your code here
if((s.size() == 0) ^ (dict.size() == 0))
return false;
if((s.size() == 0) && (dict.size() == 0))
return true;
return partitions(s, dict, 0);
}
};

2.跄跄~~ 当然是标准答案来啦  动态规划 就是循环一遍，然后每次遍历的时候再找他的子集

这道题是一种经典案例做法，需要多做几遍，熟悉下套路！

class Solution {
public:
/**
* @param s: A string s
* @param dict: A dictionary of words dict
*/
int getMaxLength(unordered_set<string> &dict) {
int maxLength = 0; // 试试看中文
for (unordered_set<string>::iterator it = dict.begin(); it != dict.end(); ++it) {
maxLength = maxLength > (*it).length() ? maxLength : (*it).length();
}
return maxLength;
}

bool wordBreak(string s, unordered_set<string> &dict) {
int maxLength = getMaxLength(dict);
bool *canSegment = new bool[s.length() + 1];
canSegment[0] = true;
for (int i = 1; i <= s.length(); i++) {
canSegment[i] = false;
for (int j = 1; j <= maxLength && j <= i; j++) {
if (!canSegment[i - j]) {
continue;
}
string word = s.substr(i - j, j); //慢慢讲切割长度拉长
if (dict.find(word) != dict.end()) {
canSegment[i] = true;
break;
}
}
}

return canSegment[s.length()];
}
};