[LeetCode] Word Break, Solution

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".

[Thoughts]

一个DP问题。定义possible[i] 为S字符串上[0,i]的子串是否可以被segmented by dictionary.

那么

possible[i] = true      if  S[0,i]在dictionary里面

                = true      if   possible[k] == true 并且 S[k+1,j]在dictionary里面， 0<k<i

               = false      if    no such k exist.

 

[Code]

实现时前面加一个dummy节点，这样可以把三种情况统一到一个表达式里面。

   
   

    
    
    
     1 
    
        
    
    bool
    
     wordBreak(
    
    string
    
     s, unordered_set
    
    <
    
    string
    
    >
    
     
    
    &
    
    dict) {

    
     2 
    
            
    
    string
    
     s2 
    
    =
    
     
    
    '
    
    #
    
    '
    
     
    
    +
    
     s;

    
     3 
    
            
    
    int
    
     len 
    
    =
    
     s2.size();

    
     4 
    
            vector
    
    <
    
    bool
    
    >
    
     possible(len, 
    
    0
    
    );

    
     5 
    
            

    
     6 
    
            possible[
    
    0
    
    ] 
    
    =
    
     
    
    true
    
    ;

    
     7 
    
            
    
    for
    
    (
    
    int
    
     i 
    
    =
    
    1
    
    ; i
    
    <
    
     len; 
    
    ++
    
    i)

    
     8 
    
            {

    
     9 
    
                
    
    for
    
    (
    
    int
    
     k
    
    =
    
    0
    
    ; k
    
    <
    
    i; 
    
    ++
    
    k)

    
    10 
    
                {

    
    11 
    
                    possible[i] 
    
    =
    
     possible[k] 
    
    &&
    
     

    
    12 
    
                                    dict.find(s2.substr(k
    
    +
    
    1
    
    , i
    
    -
    
    k)) 
    
    !=
    
     dict.end();

    
    13 
    
                    
    
    if
    
    (possible[i]) 
    
    break
    
    ;

    
    14 
    
                }

    
    15 
    
            }

    
    16 
    
            

    
    17 
    
            
    
    return
    
     possible[len
    
    -
    
    1
    
    ];

    
    18 
    
        }