【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()+, false);
         dppos[]=true;

         for(int i=;i<dppos.size();i++){
             for(int j=i-;j>=;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()];
     }
 };