【问题描述】[中等]

【解答思路】

1. 动态规划

动态规划流程
第 1 步：设计状态
dp[i] 表示字符串的前 i 个字符的最少未匹配数。

第 2 步：状态转移方程
假设当前我们已经考虑完了前 i -1个字符了，对于前 i 个字符对应的最少未匹配数：

第 i 个字符未匹配，则 dp[i] = dp[i+1] + 1，即不匹配数加 1;
遍历前 i -1个字符，若以其中某一个下标 j 为开头、以第 i 个字符为结尾的字符串正好在词典里，则 dp[i] = min(dp[ i ], dp[ j ]) 更新 dp[i]。

第 3 步：考虑初始化
int[] dp = new int[n+1];
dp[0] = 0;
第 4 步：考虑输出
dp[n];

时间复杂度：O(N^3) 空间复杂度：O(N)

class Solution {public int respace(String[] dictionary, String sentence) {Set<String> dic = new HashSet<>();for(String str: dictionary) dic.add(str);int n = sentence.length();//dp[i]表示sentence前i个字符所得结果int[] dp = new int[n+1];for(int i=1; i<=n; i++){dp[i] = dp[i-1]+1;  //先假设当前字符作为单词不在字典中for(int j=0; j<i; j++){if(dic.contains(sentence.substring(j,i))){dp[i] = Math.min(dp[i], dp[j]);}}}return dp[n];}
}

2. Trie字典树优化

复杂度分析

class Solution {public int respace(String[] dictionary, String sentence) {int n = sentence.length();Trie root = new Trie();for (String word: dictionary) {root.insert(word);}int[] dp = new int[n + 1];Arrays.fill(dp, Integer.MAX_VALUE);dp[0] = 0;for (int i = 1; i <= n; ++i) {dp[i] = dp[i - 1] + 1;Trie curPos = root;for (int j = i; j >= 1; --j) {int t = sentence.charAt(j - 1) - 'a';if (curPos.next[t] == null) {break;//单词终结标志} else if (curPos.next[t].isEnd) {dp[i] = Math.min(dp[i], dp[j - 1]);}if (dp[i] == 0) {break;}curPos = curPos.next[t];}}return dp[n];}
}class Trie {public Trie[] next;public boolean isEnd;public Trie() {next = new Trie[26];isEnd = false;}public void insert(String s) {Trie curPos = this;for (int i = s.length() - 1; i >= 0; --i) {int t = s.charAt(i) - 'a';if (curPos.next[t] == null) {curPos.next[t] = new Trie();}curPos = curPos.next[t];}//给遍历的时候的单词终结标志curPos.isEnd = true;}
}

3. 字符串哈希

复杂度分析

class Solution {static final long P = Integer.MAX_VALUE;static final long BASE = 41;public int respace(String[] dictionary, String sentence) {Set<Long> hashValues = new HashSet<Long>();for (String word : dictionary) {hashValues.add(getHash(word));}int[] f = new int[sentence.length() + 1];Arrays.fill(f, sentence.length());f[0] = 0;for (int i = 1; i <= sentence.length(); ++i) {f[i] = f[i - 1] + 1;long hashValue = 0;for (int j = i; j >= 1; --j) {int t = sentence.charAt(j - 1) - 'a' + 1;hashValue = (hashValue * BASE + t) % P;if (hashValues.contains(hashValue)) {f[i] = Math.min(f[i], f[j - 1]);}}}return f[sentence.length()];}public long getHash(String s) {long hashValue = 0;for (int i = s.length() - 1; i >= 0; --i) {hashValue = (hashValue * BASE + s.charAt(i) - 'a' + 1) % P;}return hashValue;}
}

【总结】

1.动态规划流程

第 1 步：设计状态
第 2 步：状态转移方程
第 3 步：考虑初始化
第 4 步：考虑输出
第 5 步：考虑是否可以状态压缩

2. Rabin-Karp 字符串编码 (字符串哈希)

3.Trie

class Trie {public Trie[] next;public boolean isEnd;public Trie() {next = new Trie[26];isEnd = false;}public void insert(String s) {Trie curPos = this;for (int i = s.length() - 1; i >= 0; --i) {int t = s.charAt(i) - 'a';if (curPos.next[t] == null) {curPos.next[t] = new Trie();}curPos = curPos.next[t];}curPos.isEnd = true;}
}

转载链接：https://leetcode-cn.com/problems/re-space-lcci/solution/hui-fu-kong-ge-by-leetcode-solution/
Rabin-Karp 字符串编码 参考链接：https://leetcode-cn.com/problems/longest-happy-prefix/solution/zui-chang-kuai-le-qian-zhui-by-leetcode-solution/