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Commit ab4ad45

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Bruce YangBruce Yang
Bruce Yang
authored and
Bruce Yang
committed
Merge branch 'master' of github.com:yanglr/leetcode-ac
2 parents 92c1af9 + 08b7fac commit ab4ad45

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‎algo-basic/acwing167-sticks.cpp

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#include <algorithm>
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// #include <cstring>
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#include <iostream>
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using namespace std;
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const int N = 70;
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int n; // 所有木棒的个数
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int sum, length; /* sum: 所有木棒的长度总和, length: 当前枚举的木棍的长度 */
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int sticks[N]; // 存所有的木棒
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bool st[N]; // 存每根木棒当前有没有被用过
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/**
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* u: 当前拼到了第几根(1, 2...)木棍, 木棒内部的编号No.
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cur: 当前这根木棍已经拼了多长
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start: 当前从原木棒数组的哪个index开始枚举
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*/
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bool dfs(int u, int cur, int start)
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{
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if (u * length == sum) // 找到了一组解
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{
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return true;
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}
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if (cur == length) /* 搜索下一个新的木棍 */
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return dfs(u + 1, 0, 0);
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for (int i = start; i < n; i++) // 循环需要从start开始
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{
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if (st[i]) continue; // 当前木棍用掉了
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int l = sticks[i];
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if (cur + l <= length)
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{
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st[i] = true;
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if (dfs(u, cur + l, i + 1))
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{
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return true;
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}
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st[i] = false; // 清空访问状态
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// 剪枝3 如果是第一根木棍放入失败, 则一定失败
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if (!cur)
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{
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return false;
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}
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//剪枝4 如果是最后一根木棒放入失败, 则一定失败
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if (cur + l == length)
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{
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return false;
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}
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// 剪枝2 如果失败了, 则跳过长度相等的木棒
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int j = i;
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while (j < n && sticks[j] == l)
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{
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j++;
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}
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i = j - 1;
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}
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}
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return false;
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}
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/**
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输入样例:
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9
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5 2 1 5 2 1 5 2 1
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4
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1 2 3 4
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0
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*/
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// Test
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int main()
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{
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while (cin >> n, n)
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{
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sum = 0, length = 0;
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fill(st, st + N, false); // 把所有长度>50的木棍的访问状态清空一下
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for (int i = 0; i < n; i++)
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{
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cin >> sticks[i];
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if (sticks[i] > 50)
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{
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continue;
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}
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sum += sticks[i];
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length = max(length, sticks[i]);
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}
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// 剪枝: 优化搜索顺序
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sort(sticks, sticks + n, greater<int>());
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for (int i = 0; i < n; i++)
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{
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if (sticks[i] > 50)
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{
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st[i] = true;
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}
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}
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while (true)
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{
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if (sum % length == 0 && dfs(0, 0, 0))
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{
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cout << length << endl;
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break;
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}
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length++;
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}
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}
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return 0;
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}

‎algo-basic/in167.txt

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#include<vector>
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#include<algorithm>
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#include<iostream>
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#include<unordered_map>
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using namespace std;
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class Solution {
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private:
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struct Node {
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int idx;
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vector<Node*> children;
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Node() : idx(-1), children(26, nullptr) {}
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};
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struct Trie {
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Node* root;
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Trie() : root(new Node()) {}
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void insert(string& word, int idx)
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{
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Node* p = root;
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for (char& c : word) {
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c -= 'a';
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if (p->children[c] == nullptr)
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{
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p->children[c] = new Node();
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}
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p = p->children[c];
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}
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p->idx = idx;
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}
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};
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public:
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vector<vector<int>> multiSearch(string big, vector<string>& smalls) {
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unordered_map<string, vector<int>> cache;
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const int n = big.size();
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const int m = smalls.size();
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vector<vector<int>> res(m);
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Trie trie = Trie();
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// 构造前缀树
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for (int i = 0; i < m; i++)
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{
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trie.insert(smalls[i], i);
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}
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for (int i = 0; i < n; i++)
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{
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int j = i;
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Node* node = trie.root;
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while (j < n && node->children[big[j] - 'a'])
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{
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node = node->children[big[j] - 'a'];
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if (node->idx != -1)
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{
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res[node->idx].push_back(i);
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}
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j++;
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}
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}
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return res;
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}
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};
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// Test
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int main()
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{
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Solution sol;
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string big = "mississippi";
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vector<string> smalls = {"is", "ppi", "hi", "sis", "i", "ssippi"};
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auto res = sol.multiSearch(big, smalls);
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for (auto& row : res) // 遍历每一行
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{
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for (auto& elem : row) // 输出每一个元素
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cout << elem << " ";
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cout << "\n";
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}
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return 0;
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}
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#include<vector>
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#include<algorithm>
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#include<iostream>
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#include<cstring>
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using namespace std;
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class Solution {
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public:
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vector<string> commonChars(vector<string>& words) {
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vector<string> res;
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int minDict[26]; // 统计每个字母的最小count
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fill(minDict, minDict + 26, INT_MAX);
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for (int j = 0; j < words.size(); j++) // 将每一个单词进行哈希化, 然后分别算出每个字母的最小count
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{
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string curWord = words[j];
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vector<int> curDict(26, 0); // 26个字母
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for (auto& ch : curWord)
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curDict[ch - 'a']++;
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for (int i = 0; i < 26; i++)
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minDict[i] = min(minDict[i], curDict[i]);
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}
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for (int i = 0; i < 26; i++)
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{
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int curCount = minDict[i];
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if (curCount >= 1)
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{
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auto ch = i + 'a';
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string str;
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str.push_back(ch);
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for (int i = 0; i < curCount; i++)
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res.push_back(str);
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}
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}
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return res;
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}
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};
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// Test
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int main()
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{
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Solution sol;
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vector<string> words = {"cool", "lock", "cook"};
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auto res = sol.commonChars(words);
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for (auto str : res)
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cout << str << endl;
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return 0;
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}
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#include<vector>
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#include<algorithm>
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#include<iostream>
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using namespace std;
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class Solution {
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public:
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bool isValid(string s) {
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string stk; // 用string模拟栈
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for (auto& c: s) {
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stk += c;
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if (stk.size() >= 3 && stk.substr(stk.size() - 3) == "abc")
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{
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for (int i = 0; i < 3; i++)
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stk.pop_back();
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}
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}
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return stk.empty();
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}
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};
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// Test
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int main()
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{
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Solution sol;
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string s = "abcabcababcc";
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auto res = sol.isValid(s);
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cout << (res ? "True" : "False") << endl;
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return 0;
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}
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#include<vector>
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#include<algorithm>
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#include<iostream>
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using namespace std;
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class Solution {
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public:
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int longestOnes(vector<int>& nums, int k) {
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const int N = nums.size();
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int i = 0; // i: 滑动窗口左边界, 固定右边界, 移动左边界
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int count = 0; // 已经使用了多少次反转次数
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int res = 0; // res: max len
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for (int j = 0; j < N; j++)
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{
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if (nums[j] == 1) // 新加入窗口右侧的数是1, 不需要使用反转次数
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res = max(res, j - i + 1);
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else {
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count++;
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while (count > k) // 此时使用的反转次数超了, 需要移动左边界, 直到窗口使用的反转次数减小回到k为止
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{
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if (nums[i] == 0)
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count--;
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i++;
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}
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res = max(res, j - i + 1);
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}
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}
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return res;
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}
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};
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// Test
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int main()
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{
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Solution sol;
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vector<int> nums = {1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0};
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int k = 2;
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auto res = sol.longestOnes(nums, k);
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cout << res << endl;
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return 0;
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}
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#include<vector>
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#include<algorithm>
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#include<iostream>
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using namespace std;
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class Solution {
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public:
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int longestOnes(vector<int>& nums, int k) {
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const int N = nums.size();
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int left = 0; // left: 滑动窗口左边界, 固定右边界, 移动左边界
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int j = 0; // j: 滑动窗口右边界right
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int preSum = 0; // 使用的反转次数 0 -> 1
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int res = 0;
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for (int j = 0; j < N; j++)
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{
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preSum += (nums[j] == 0 ? 1 : 0);
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if (preSum > k) // 一开始, 让反转次数达到k+1, 然后开始根据情况收缩左边界
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{
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while (left < j && nums[left] == 1) /* 在左边界每遇到一个1, 就直接移动左边界, 如果遇到1个0, 翻转次数又新腾出了1, 需要先更新preSum的值 */
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left++;
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preSum -= 1;
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left++;
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}
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res = max(res, j - left + 1); // 此时, 反转次数恰好恢复到k, 可以更新结果了
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}
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return res;
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}
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};
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// Test
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int main()
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{
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Solution sol;
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vector<int> nums = {1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0};
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int k = 2;
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auto res = sol.longestOnes(nums, k);
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cout << res << endl;
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return 0;
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}

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