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Update 0139.单词拆分.md
删除原Line 102,修改memory数组为bool型 因为根据执行顺序,Line 101的if判断句,只有在前一个判断返回true的时候才会递归,因此若执行到memory[startIndex] = 1时,程序已经完成了遍历,memory[startIndex] = 1的情况完全没用的上。而memory用上的情况为false重复,即程序已经判断过startIndex开头无法分割。
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@ -89,27 +89,26 @@ class Solution {
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private:
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bool backtracking (const string& s,
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const unordered_set<string>& wordSet,
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vector<int>& memory,
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vector<bool>& memory,
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int startIndex) {
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if (startIndex >= s.size()) {
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return true;
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}
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// 如果memory[startIndex]不是初始值了,直接使用memory[startIndex]的结果
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if (memory[startIndex] != -1) return memory[startIndex];
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if (!memory[startIndex]) return memory[startIndex];
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for (int i = startIndex; i < s.size(); i++) {
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string word = s.substr(startIndex, i - startIndex + 1);
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if (wordSet.find(word) != wordSet.end() && backtracking(s, wordSet, memory, i + 1)) {
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memory[startIndex] = 1; // 记录以startIndex开始的子串是可以被拆分的
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return true;
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}
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}
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memory[startIndex] = 0; // 记录以startIndex开始的子串是不可以被拆分的
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memory[startIndex] = false; // 记录以startIndex开始的子串是不可以被拆分的
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return false;
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}
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public:
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bool wordBreak(string s, vector<string>& wordDict) {
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unordered_set<string> wordSet(wordDict.begin(), wordDict.end());
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vector<int> memory(s.size(), -1); // -1 表示初始化状态
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vector<bool> memory(s.size(), 1); // -1 表示初始化状态
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return backtracking(s, wordSet, memory, 0);
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}
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};
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