Leetcode-word break

Given A string s and a Dictionary of words dict, determine if s can segmented into a space-s Eparated 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" .

A look at this topic, it feels inside backtracking, because this problem, in the backtracking method in the usual. On the code:

public Boolean wordbreak (String s, set<string> dict) {    StringBuilder sb = new StringBuilder ();        Return helper (S, dict, 0, SB);    }    Private Boolean helper (String s, set<string> dict, int index, StringBuilder sb) {    if (index = = dict.size ()) {
   return s.equals (sb.tostring ());    }    Iterator<string> iter = Dict.iterator ();    while (Iter.hasnext ()) {    StringBuilder sb1 = new StringBuilder (SB);    String dic = Iter.next ();    Sb.append (DIC);    Iter.remove ();    if (helper (s, Dict, Index+1, SB))    return true;    Dict.add (DIC);    SB = SB1;    }    return false;    }

Result timed out! In general, the time limit for backtracking is relatively wide, but timed out! There must be a space for optimization, pruning AH:,

public Boolean wordbreak (String s, set<string> dict) {    StringBuilder sb = new StringBuilder ();        Return helper (S, dict, 0, SB);    }    Private Boolean helper (String s, set<string> dict, int index, StringBuilder sb) {    if (!s.startswith (sb.tostring ())) {    return false;    }    if (index = = dict.size ()) {    return s.equals (sb.tostring ());    }    Iterator<string> iter = Dict.iterator ();    while (Iter.hasnext ()) {    StringBuilder sb1 = new StringBuilder (SB);    String dic = Iter.next ();    Sb.append (DIC);    Iter.remove ();    if (helper (s, Dict, Index+1, SB))    return true;    Dict.add (DIC);    SB = SB1;    }    return false;    }

In fact, this pruning can subtract many branches, but still timeout! What to do? Only with the most efficient dynamic planning, on the code:

public Boolean WORDBREAK1 (String s, set<string> dict) {    int length = S.length ();        Boolean[] can = new boolean[length+1];        Can[0] = true;        for (int i = 1, i <= length; i++) {for            (int j = 0; J < i; J + +) {                if (Can[j] && dict.contains (s.subs Tring (J, i))) {                    Can[i] = true;                    Break        ;        }}} return can[length];    }

A Boolean array of can[i] means that s.substring (0,i) is divisible, and the last element of the array is the one you are seeking.

Leetcode-word break