LeetCode OJ 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".

DP:

first dp:

class Solution {public:    bool wordBreak(string s, unordered_set<string> &dict) {        vector<short> isOK(s.size() + 1, 0);        isOK[s.size()] = 1;        for (int i = s.size() - 1; i >= 0; i--) {            for (int j = i; j < s.size(); j++) {                if (dict.find(s.substr(i, j - i + 1)) != dict.end() && isOK[j + 1]) {                    isOK[i] = 1;                    break;                }            }        }        return isOK[0];    }};

The thought is :

This dp is from the bottom. Suppose the string is "abcdefghijk", the word break is successful when "abcde" is in the dictionary and "fghijk" is successfully word break.

So, isOK[i] means the substr starting from i is successfully word break. In the inner loop, if we found a substr from i to j is in the dictionary and the rest substr from j + 1 to the end is OK, then isOK[i] = 1.

Notice that isOK[s.size()] = 1 because the substr is empty.

second dp:

class Solution {public:    bool wordBreak(string s, unordered_set<string> &dict) {        vector<short> isOK(s.size() + 1, 0);        isOK[0] = 1;        for (int i = 1; i <= s.size(); i++) {            for (int j = 1; j <= i; j++) {                if (dict.find(s.substr(j - 1, i - j + 1)) != dict.end() && isOK[j - 1]) {                    isOK[i] = 1;                    break;                }            }        }        return isOK[s.size()];    }};

Notice that isOK[0] is just a start sign of 0 length, which is always true, so the i and j start from 1 but they both point to s[i - 1] or s[j - 1].