【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"

.

Solution:

dppos[i]==true/false表示字符串从开头到i的子串是否存在cut方案满足条件

动态规划设置初值bpos[0]==true

string.substr(int beginIndex, int length): 取string从beginIndex开始长length的子串

class Solution {
public:
bool wordBreak(string s, unordered_set<string>& wordDict) {　　　　//runtime:4ms
vector<bool> dppos(s.size()+1, false);
dppos[0]=true;

for(int i=1;i<dppos.size();i++){
for(int j=i-1;j>=0;j--){　　//从右到左找快很多

if(dppos[j]==true && wordDict.find(s.substr(j,i-j))!=wordDict.end()){　　
dppos[i]=true;
break;　　　//只要找到一种切分方式就说明长度为i的单词可以成功切分，因此可以跳出内层循环
}
}
}
return dppos[s.size()];
}
};