[leetcode 139]Word Break

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"

.

对从第一个字符到最后一个字符进行切分，新建一个数组备忘

count[i]=1表示字符串s从第一个字符到弟i个字符可以由字典组成

遍历字符串不断切分，返回count[s.size()]

AC代码

class Solution {
public:
bool wordBreak(string s, unordered_set<string>& wordDict) {
int len=s.size();
int sum=wordDict.size();
if(len==0&&sum==0)
return true;
else if(len==0||sum==0)
return false;
int count[len+1];
memset(count,0,sizeof(count));
count[0]=1;
string temp="";
for(int i=1;i<=len;++i)
{
for(int j=i-1;j>=0;--j)
{
temp=s.substr(j,i-j);
if(count[j]&&wordDict.find(temp)!=wordDict.end())
{
count[i]=1;
break;
}
}
}
return count[len];
}
};

其他Leetcode题目AC代码：https://github.com/PoughER/leetcode