[LeetCode]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"

.

Analysis:

DP problem. Define possible[i]: substring of s 0..i could be segmented in dict

possible[i] = true if s[0.i] in dict

= true if possible[k]=true && s[k+1,i] in dict (0<k<i)

= false other situations

java

public boolean wordBreak(String s, Set<String> dict) {
		int len = s.length();
		boolean []possible = new boolean[len+1];
		possible[0] = true;
		for(int i=1;i<=len;i++){
			for(int j=0;j<i;j++){
				if(possible[j]== true && dict.contains(s.substring(j,i))){
					possible[i] = true;
					break;
				}
			}
		}
		return possible[len];
    }

c++

bool wordBreak(string s, unordered_set<string> &dict) {
        string s2 = '#'+s;
    int len = s2.size();
    vector<bool> possible(len,0);
    possible[0] = true;
    for(int i=1;i<len;i++){
        for(int k=0;k<i;k++){
            possible[i] = possible[k] && dict.find(s2.substr(k+1,i-k))!=dict.end();
            if(possible[i]) break;
        }
    }
    return possible[len-1];
    }

这里可以不用加#的，参照java代码

public String substring(int beginIndex,
               int endIndex)

Returns a new string that is a substring of this string. The substring begins at the specified

beginIndex

and extends to the character at index

endIndex - 1

. Thus the length of the substring is

endIndex-beginIndex

.
Examples:

"hamburger".substring(4, 8) returns "urge"
 "smiles".substring(1, 5) returns "mile"

Parameters:

beginIndex

- the beginning index, inclusive.

endIndex

- the ending index, exclusive. Returns:the specified substring.