[LeetCode] Word Break 拆分词句

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".

这道拆分词句问题是看给定的词句能分被拆分成字典里面的内容，还是需要用动态规划Dynamic Programming来做，具体讲解可参考网友Code Ganker的博客，代码如下：

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

下面我们从题目中给的例子来分析：

le e

lee ee e

leet

leetc eetc etc tc c

leetco eetco etco tco co o

leetcod eetcod etcod tcod cod od d

leetcode eetcode etcode tcode code

T F F F T F F F T

我们知道算法的核心思想是逐行扫描，每一行再逐个字符扫描，每次都在组合出一个新的字符串都要到字典里去找，如果有的话，则跳过此行，继续扫描下一行。

时间： 2024-08-08 05:21:35

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