[Leetcode Week9]Word Break

最新推荐文章于 2020-07-16 16:10:05 发布

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最新推荐文章于 2020-07-16 16:10:05 发布

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原文链接：http://www.cnblogs.com/yanhewu/p/7768078.html

本文介绍了一种利用动态规划解决 LeetCode 上 WordBreak 问题的方法。该方法通过生成源字符串的所有子串并在词典中查找这些子串来判断原始字符串是否能被分割为词典中的词汇。

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Word Break 题解

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题目来源：https://leetcode.com/problems/word-break/description/

Description

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".

Solution

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