【C++】【LeetCode】139. Word Break

题目

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. You may assume the dictionary does not contain duplicate words.

For example, given
s = “leetcode”,
dict = [“leet”, “code”].

Return true because “leetcode” can be segmented as “leet code”.

思路

建一个数组用于存到第i个位置为止能不能用wordDict里的词表示出来，能为true，不能为false。当遍历到第i个位置时，检查在它之前的位置j为true或者false，如果为true，则检查从i到j的字符串是不是在wordDict中。

代码

class Solution {public:    bool wordBreak(string s, vector<string>& wordDict) {        if (wordDict.size() == 0) {            return false;        }        vector<bool> endHere(s.size() + 1, false);        endHere[0] = true;        for (int i = 1; i <= s.size(); i++) {            for (int j = i - 1; j >= 0; j--) {                if (endHere[j]) {                    string word = s.substr(j, i - j);                    if (find(wordDict.begin(), wordDict.end(), word) != wordDict.end()) {                        endHere[i] = true;                        break;                    }                }            }        }        return endHere[s.size()];    }};