Leetcode-139. Word Break

题目：

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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AC代码1：
class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict) {
int size = wordDict.size();
set<string>ma;
for (vector<string> ::iterator iter = wordDict.begin(); iter != wordDict.end(); iter++)ma.insert((*iter));
int len = s.size();
vector<bool>re(len, false);
string temp;
for (int i = 0; i < len; i++){
for (int j = i; j >= 0; j--){
if (re[j] == true){
temp = s.substr(j + 1, i - j);
if (ma.find(temp) != ma.end()){
re[i] = true;
break;
}
}
if (j == 0){
temp = s.substr(0, i + 1);
if (ma.find(temp) != ma.end()){
re[i] = true;
break;
}
}
}
}
return re[len - 1];
}
};

AC代码2：
class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict) {
int size = wordDict.size();
set<string>ma;
for (vector<string> ::iterator iter= wordDict.begin();iter!=wordDict.end(); iter++)ma.insert((*iter));
int len = s.size();
vector<bool>re(len+1,false);
re[0] = true;
string temp;
for (int i = 1; i <=len;i++){
for (int j = i-1; j >= 0;j--){
if (re[j] == true){
temp = s.substr(j, i - j);
if (ma.find(temp) != ma.end()){
re[i] = true;
break;
}
}
}
}
return re[len];
}
};

解析：
AC代码1是自己写的第一个版本，这时需要注意的的一种情况就是当前下标为i的字母要包含下标为0的字母所组成的字符串存在于wordDict中；

AC代码2利用一个额外的存储空间并将re[0]设置为true，这样的好处就是将AC代码1中的特殊情况统一处理，也就是当j递减到0时，那么这时字符串s就会将第

一个字母包含就去，这种统一化处理方式比较妙！