[LeetCode] Word Break

最新推荐文章于 2023-06-01 11:39:37 发布

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本文介绍了一种使用动态规划解决字符串能否通过字典中的词进行有效切割的问题。给出的算法通过构建二维数组来记录子串的有效切割状态，并通过迭代填充该数组以最终确定整个字符串是否能被成功切割。

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".

一个字符串是否能够由字典中的词切割

动规求解

f[i][j] = 1表示s[i-j]可以被dict切割

那么若f[i][j] == 1 且 f[j+1][k] == 1 则有f[i][k] = 1

class Solution {
public:
    bool wordBreak(string s, unordered_set<string> &dict) {
        // Note: The Solution object is instantiated only once and is reused by each test case
        if(s.length() == 0) return true;
        
        int len = s.length();
        int ** f = new int*[len];
        for(int i = 0; i < len; i++) {
            f[i] = new int[len];
            for(int j = 0; j < len; j++) {
                string tmp = s.substr(i,j-i+1);
                if(dict.find(tmp) != dict.end()) f[i][j] = 1;
                else f[i][j] = 0;
            }
        }
        
        for(int i = 0; i < len; i++) {
            for(int j = i; j < len; j++) {
                for(int k = j+1; k < len; k++) {
                    if(f[i][j] == 1 && f[j+1][k] == 1) {
                        f[i][k] = 1;
                    }
                }
            }
        }
        return f[0][len-1] == 1;
    }
};