|
| 1 | + |
| 2 | +// 如何往 完全背包上靠？ |
| 3 | +// 用多次倒是可以往 完全背包上靠一靠 |
| 4 | +// 和单词分割的问题有点像 |
| 5 | + |
| 6 | +[回溯算法：分割回文串](https://mp.weixin.qq.com/s/Pb1epUTbU8fHIht-g_MS5Q) |
| 7 | + |
| 8 | +回溯法代码： |
| 9 | +``` |
| 10 | +class Solution { |
| 11 | +private: |
| 12 | + bool backtracking (const string& s, const unordered_set<string>& wordSet, int startIndex) { |
| 13 | + if (startIndex >= s.size()) { |
| 14 | + return true; |
| 15 | + } |
| 16 | + for (int i = startIndex; i < s.size(); i++) { |
| 17 | + string word = s.substr(startIndex, i - startIndex + 1); |
| 18 | + if (wordSet.find(word) != wordSet.end() && backtracking(s, wordSet, i + 1)) { |
| 19 | + return true; |
| 20 | + } |
| 21 | + } |
| 22 | + return false; |
| 23 | + } |
| 24 | +public: |
| 25 | + bool wordBreak(string s, vector<string>& wordDict) { |
| 26 | + unordered_set<string> wordSet(wordDict.begin(), wordDict.end()); |
| 27 | + return backtracking(s, wordSet, 0); |
| 28 | + } |
| 29 | +}; |
| 30 | +``` |
| 31 | + |
| 32 | +``` |
| 33 | +"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab" |
| 34 | +["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"] |
| 35 | +``` |
| 36 | + |
| 37 | +可以使用一个一维数组保存一下，递归过程中计算的结果，C++代码如下： |
| 38 | + |
| 39 | +使用memory数组保存 每次计算的以startIndex起始的计算结果，如果memory[startIndex]里已经被赋值了，直接用memory[startIndex]的结果。 |
| 40 | +``` |
| 41 | +class Solution { |
| 42 | +private: |
| 43 | + bool backtracking (const string& s, |
| 44 | + const unordered_set<string>& wordSet, |
| 45 | + vector<int>& memory, |
| 46 | + int startIndex) { |
| 47 | + if (startIndex >= s.size()) { |
| 48 | + return true; |
| 49 | + } |
| 50 | + // 如果memory[startIndex]不是初始值了，直接使用memory[startIndex]的结果 |
| 51 | + if (memory[startIndex] != -1) return memory[startIndex]; |
| 52 | + for (int i = startIndex; i < s.size(); i++) { |
| 53 | + string word = s.substr(startIndex, i - startIndex + 1); |
| 54 | + if (wordSet.find(word) != wordSet.end() && backtracking(s, wordSet, memory, i + 1)) { |
| 55 | + memory[startIndex] = 1; // 记录以startIndex开始的子串是可以被拆分的 |
| 56 | + return true; |
| 57 | + } |
| 58 | + } |
| 59 | + memory[startIndex] = 0; // 记录以startIndex开始的子串是不可以被拆分的 |
| 60 | + return false; |
| 61 | + } |
| 62 | +public: |
| 63 | + bool wordBreak(string s, vector<string>& wordDict) { |
| 64 | + unordered_set<string> wordSet(wordDict.begin(), wordDict.end()); |
| 65 | + vector<int> memory(s.size(), -1); // -1 表示初始化状态 |
| 66 | + return backtracking(s, wordSet, memory, 0); |
| 67 | + } |
| 68 | +}; |
| 69 | +``` |
| 70 | + |
| 71 | + |
| 72 | +得好好分析一下，完全背包和01背包，这个对于刷leetcode太重要了 |
| 73 | + |
| 74 | +注意这里要空出一个 dp[0] 来做起始位置 |
| 75 | +``` |
| 76 | +class Solution { |
| 77 | +public: |
| 78 | + bool wordBreak(string s, vector<string>& wordDict) { |
| 79 | + unordered_set<string> wordSet(wordDict.begin(), wordDict.end()); |
| 80 | + vector<bool> dp(s.size() + 1, false); |
| 81 | + dp[0] = true; |
| 82 | + for (int i = 1; i <= s.size(); i++) { |
| 83 | + for (int j = 0; j < i; j++) { |
| 84 | + string word = s.substr(j, i - j); //substr(起始位置，截取的个数) |
| 85 | + if (wordSet.find(word) != wordSet.end() && dp[j]) { |
| 86 | + dp[i] = true; |
| 87 | + } |
| 88 | + } |
| 89 | + //for (int k = 0; k <=i; k++) cout << dp[k] << " "; |
| 90 | + //cout << endl; |
| 91 | + } |
| 92 | + return dp[s.size()]; |
| 93 | + } |
| 94 | +}; |
| 95 | +``` |
| 96 | +时间复杂度起始是O(n^3)，因为substr返回子串的副本是O(n)的复杂度（n是substring的长度） |
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