leetcode:word-break
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1.题目描述
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".
2.题目分析分析可得可以用动态规划,如:leetcode:
若l和eetcode在字典里则是;
若le和etcode在字典里则是;
若lee和tcode在字典里则是;
若leet和code在字典里则是;
若leetc和ode在字典里则是;
若leetco和de在字典里则是;
若leetcod和e在字典里则是;
只要找到一种情况leetcode就是可以的。
状态转移方程:
* f(i) 表示s[0,i]是否可以分词
* f(i) = f(j) && f(j+1,i); 0 <= j < i;
string substr(int pos = 0,int n = npos) const;//返回pos开始的n个字符组成的字符串
3.代码实现
class Solution {public: bool wordBreak(string s, unordered_set<string> &dict) { int length=s.length(); vector<bool> array(length+1,false); array[0]=true; for(int j=1;j<=length;j++) { for(int i=0;i<j;i++) { if(array[i] && dict.find(s.substr(i,j-i))!=dict.end()) { array[j]=true; break; } } } return array[length]; }};
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