LeetCode: Word Break-优快云博客

本文介绍了一种使用动态规划解决字符串拆分问题的方法，通过给定的字典判断一个字符串是否能被拆分为一个或多个字典中的单词。文章提供了两种实现方式，并对第二种方法进行了优化，减少不必要的子串检查。

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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”.

思路是DP。dp[i]表示前i个字符可以被word break。所以可以得到状态转移方程：dp[i] = dp[j] && dict.contains(s[j ~ i])

public class Solution {
    public boolean wordBreak(String s, Set<String> dict) {
        StringBuilder sb = new StringBuilder();
        boolean[] can = new boolean[s.length() + 1];
        can[0] = true;
        for(int i = 1; i < can.length; i++){
            for(int j = 0; j < i; j++){
                can[i] |= can[j] && dict.contains(s.substring(j, i));
            }
        }
        return can[s.length()];
    }
}

这里还可以预处理优化下，先O(n)扫一遍得出字典中最短和最长单词的长度。每次分割只选取[min, max]长度之间的substring。

public class Solution {
    public boolean wordBreak(String s, Set<String> dict) {
        int min = Integer.MAX_VALUE;
        int max = Integer.MIN_VALUE;
        for(String str : dict){
            min = Math.min(str.length(), min);
            max = Math.max(str.length(), max);
        }
        boolean[] dp = new boolean[s.length() + 1];
        dp[0] = true;
        for(int i = 0; i <= s.length() - min; i++){
            for(int k = min; k <= max && i + k <= s.length(); k++){
                if(dp[i + k] || !dp[i]){
                    continue;
                }
                dp[i + k] = dp[i] && dict.contains(s.substring(i, i + k));
            }
        }
        return dp[s.length()];
    }
}