LeetCode题解–137. Word Break

链接

LeetCode题目：https://leetcode.com/problems/word-break/

难度：Medium

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”.

题目大意是给一个字符串s和一个字典，判断s是否可以分割成一个或多个字典中的单词。

分析

设状态为match[i, j]，表示s[i, j]能否分词，则显然状态转移方程为match[i, j] = （s[i, j] ∈ dict）or （match[i, k] and match[k+1, j]),其中i<=k<=j-1，时间复杂度是O(n^3)。

代码

#include <iostream>#include <unordered_set>#include <vector>using namespace std;class Solution {public:    bool wordBreak(string s, vector<string> &wordDict) {        unordered_set<string> wordSet;        for (string word:wordDict) {            wordSet.insert(word);        }        int s_len = (int) s.size();        bool match[500][500] = {false};        for (int len = 1; len <= s_len; len++) {            for (int i = 0; i + len <= s_len; i++) {                int j = i + len - 1;                string temp = s.substr(i, len);                if (findWord(temp, wordSet)) {                    match[i][j] = true;                    continue;                }                for (int k = i; k + 1 <= j; k++) {                    if (match[i][k] && match[k + 1][j]) {                        match[i][j] = true;                        continue;                    }                }            }        }        return match[0][s_len - 1];    }    bool findWord(string word, unordered_set<string> &wordSet) {        return wordSet.find(word) != wordSet.end();    }};