20-01. 找到需要补充粉笔的学生编号

class Solution {public:int chalkReplacer(vector<int> &chalk, int k) {int n = chalk.size();if (chalk[0] > k) return 0;for (int i = 1; i < n; ++i) {chalk[i] += chalk[i - 1];if (chalk[i] > k) return i;}k %= chalk.back();return upper_bound(chalk.begin(), chalk.end(), k) - chalk.begin();}
};

20-02. 最小平均差

class Solution {public:int minimumAverageDifference(const vector<int> &nums) {int n = nums.size();vector<long long> sum_left(n), sum_right(n, 0);vector<long long> ave_left(n), ave_right(n, 0);for (int i = 0; i < n; ++i) {sum_left[i] = (i == 0) ? nums[i] : nums[i] + sum_left[i - 1];ave_left[i] = sum_left[i] / (i + 1);}for (int i = n - 2; i >= 0; --i) {sum_right[i] = nums[i + 1] + sum_right[i + 1];ave_right[i] = sum_right[i] / (n - i - 1);}int ans = -1, diff = INT_MAX;for (int i = 0; i < n; ++i) {if (abs(ave_left[i] - ave_right[i]) < diff) {diff = abs(ave_left[i] - ave_right[i]);ans = i;}}return ans;}
};

20-03. 满足三条件之一需改变的最少字符数

class Solution {public:int minCharacters(const string &s1, const string &s2) {int n1 = s1.length(), n2 = s2.length(), ans = INT_MAX;int c1[26], c2[26];memset(c1, 0, sizeof c1);memset(c2, 0, sizeof c2);for (char c:s1) c1[c - 'a']++;for (char c:s2) c2[c - 'a']++;for (int i = 0; i < 26 && ans != 0; ++i) {ans = min(ans, n1 + n2 - c1[i] - c2[i]);if (i == 0) continue;int r1 = accumulate(c1 + i, c1 + 26, 0) + accumulate(c2, c2 + i, 0);int r2 = accumulate(c2 + i, c2 + 26, 0) + accumulate(c1, c1 + i, 0);ans = min({ans, r1, r2});}return ans;}
};

20-04. 蜡烛之间的盘子

class Solution {public:vector<int> platesBetweenCandles(string s, vector<vector<int>> &queries) {int n = s.length();vector<int> preSum(n), left(n), right(n);for (int i = 0, sum = 0; i < n; i++) {if (s[i] == '*') sum++;preSum[i] = sum;}for (int i = 0, l = -1; i < n; i++) {if (s[i] == '|') l = i;left[i] = l;}for (int i = n - 1, r = -1; i >= 0; i--) {if (s[i] == '|') r = i;right[i] = r;}vector<int> ans;for (auto &query : queries) {int x = right[query[0]], y = left[query[1]];ans.push_back(x == -1 || y == -1 || x >= y ? 0 : preSum[y] - preSum[x]);}return ans;}
};

21-01. 快照数组

class SnapshotArray {private:vector<vector<pair<int, int>>> store;int time;public:SnapshotArray(int length) : time(0) {store.assign(length, {{0, 0}});}void set(int index, int val) {int s = store[index].back().first;if (s == time) {store[index].back().second = val;} else {store[index].emplace_back(time, val);}}int snap() { return time++; }int get(int index, int snap_id) {auto &elem = store[index];int lo = 0, hi = elem.size();while (lo < hi) {int mi = lo + ((hi - lo) >> 1);elem[mi].first <= snap_id ? lo = mi + 1 : hi = mi;}return elem[--lo].second;}
};

21-02. 满足条件的子序列数目

class Solution {public:static constexpr int mod = 1e9 + 7;static constexpr int MAX_N = 1e5 + 5;int f[MAX_N];void pretreatment() {f[0] = 1;for (int i = 1; i < MAX_N; ++i) {f[i] = (long long) f[i - 1] * 2 % mod;}}int numSubseq(vector<int> &nums, int target) {pretreatment();sort(nums.begin(), nums.end());int ans = 0;for (int i = 0; i < nums.size() && nums[i] * 2 <= target; ++i) {int maxValue = target - nums[i];int pos = upper_bound(nums.begin(), nums.end(), maxValue) - nums.begin() - 1;ans = (ans + f[pos - i]) % mod;}return ans;}
};

20-03. 有界数组中指定下标处的最大值

class Solution {private:using ull = unsigned long long;ull check(ull h, ull l1, ull l2) {auto get = [](ull h, ull l) -> ull {l = min(h, l);return ((h - 1) + (h - l)) * l / 2;};return get(h, l1) + get(h, l2) + h;}public:int maxValue(int n, int index, int s) {unsigned int lo = 0, hi = s + 1, l1 = index, l2 = n - index - 1;while (lo < hi) {unsigned int mi = lo + ((hi - lo) >> 1);check(mi, l1, l2) <= s - n ? lo = mi + 1 : hi = mi;}return lo;}
};

21-01. 两个有序数组的第 K 小乘积

class Solution {private:using ll = long long;ll get(ll val, const vector<int> &nums1, const vector<int> &nums2) {ll ans = 0;for (ll x : nums1) {int lo = 0, hi = nums2.size();if (x >= 0) {while (lo < hi) {int mi = (lo + hi) >> 1;x * nums2[mi] <= val ? lo = mi + 1 : hi = mi;}ans += lo;} else {while (lo < hi) {int mi = (lo + hi) >> 1;x * nums2[mi] > val ? lo = mi + 1 : hi = mi;}ans += nums2.size() - lo;}}return ans;}public:ll kthSmallestProduct(const vector<int> &nums1, const vector<int> &nums2, ll k) {ll lo = -1e10, hi = 1e10;while (lo < hi) {ll mi = (lo + hi) >> 1;get(mi, nums1, nums2) < k ? lo = mi + 1 : hi = mi;}return lo;}
};

21-02. 设计一个文本编辑器

struct Node {char val;Node *prev, *next;Node(char val = 0) : val(val), prev(nullptr), next(nullptr) {}Node(char val, Node *prev, Node *next) : val(val), prev(prev), next(next) {}
};class TextEditor {private:Node *dummyHead, *dummyTail, *curNode;inline void insertBefore(char val) {auto node = new Node(val, curNode->prev, curNode);curNode->prev->next = node;curNode->prev = node;}inline bool deleteBefore() {if (curNode->prev == dummyHead) return false;curNode->prev = curNode->prev->prev;curNode->prev->next = curNode;return true;}string readBefore(int n = 10) {auto node = curNode;while (n-- && node->prev != dummyHead)node = node->prev;string s;while (node != curNode) {s.push_back(node->val);node = node->next;}return s;}public:TextEditor() {dummyHead = new Node();dummyTail = new Node();dummyHead->next = dummyTail;dummyTail->prev = dummyHead;curNode = dummyTail;}void addText(string text) { for (char ch:text) insertBefore(ch); }int deleteText(int k) {for (int i = 0; i < k; i++) {if (!deleteBefore()) return i;}return k;}string cursorLeft(int k) {while (k-- && curNode->prev != dummyHead)curNode = curNode->prev;return move(readBefore());}string cursorRight(int k) {while (k-- && curNode->next != nullptr) curNode = curNode->next;return move(readBefore());}
};

21-03. 翻转等价二叉树

class Solution {public:bool flipEquiv(TreeNode *root1, TreeNode *root2) {if (!root1 && !root2) return true;if (root1 && !root2) return false;if (!root1 && root2) return false;if (root1->val != root2->val) return false;return (flipEquiv(root1->left, root2->left) && flipEquiv(root1->right, root2->right)) ||(flipEquiv(root1->left, root2->right) && flipEquiv(root1->right, root2->left));}
};

21-04. 找到所有的农场组

class Solution {private:int n, m;vector<vector<int>> result;void dfs(vector<vector<int>> &land, int x, int y) {land[x][y] = 0;result.back()[2] = max(result.back()[2], x);result.back()[3] = max(result.back()[3], y);if (x + 1 < n && land[x + 1][y] == 1) dfs(land, x + 1, y);if (y + 1 < m && land[x][y + 1] == 1) dfs(land, x, y + 1);}public:vector<vector<int>> findFarmland(vector<vector<int>> &land) {n = land.size();m = land[0].size();for (int i = 0; i < n; ++i) {for (int j = 0; j < m; ++j) {if (land[i][j] == 0) continue;result.push_back({i, j, i, j});dfs(land, i, j);}}return result;}
};

21-05. 寻找重复的子树

class Solution {public:string dfs(TreeNode *root, vector<TreeNode *> &res, unordered_map<string, int> &mp) {if (!root) return "";string str = to_string(root->val) + "," + dfs(root->left, res, mp) + "," + dfs(root->right, res, mp);if (mp[str]++ == 1) res.push_back(root);return str;}vector<TreeNode *> findDuplicateSubtrees(TreeNode *root) {vector<TreeNode *> res;unordered_map<string, int> mp;dfs(root, res, mp);return res;}
};

21-06. 二叉搜索树的范围和

class Solution {public:int rangeSumBST(TreeNode *root, int low, int high) {stack<TreeNode *> st;int ans = 0;TreeNode *cur = root;while (cur || !st.empty()) {if (cur) {st.push(cur);cur = cur->left;} else {cur = st.top();st.pop();if (cur->val > high) return ans;if (cur->val >= low) ans += cur->val;cur = cur->right;}}return ans;}
};

21-07. 所有大于等于节点的值之和

class Solution {public:TreeNode *convertBST(TreeNode *root) {stack<TreeNode *> st;int ans = 0;TreeNode *cur = root;while (cur || !st.empty()) {if (cur) {st.push(cur);cur = cur->right;} else {cur = st.top();st.pop();cur->val += ans;ans = cur->val;cur = cur->left;}}return root;}
};

21-08. 把二叉搜索树转换为累加树

class Solution {public:TreeNode *convertBST(TreeNode *root) {stack<TreeNode *> st;int ans = 0;TreeNode *cur = root;while (cur || !st.empty()) {if (cur) {st.push(cur);cur = cur->right;} else {cur = st.top();st.pop();cur->val += ans;ans = cur->val;cur = cur->left;}}return root;}
};

21-09. 点菜展示表

class Solution {using foodMap = unordered_map<string, int>; // food - count
public:vector<vector<string>> displayTable(vector<vector<string>> &orders) {map<int, foodMap> tableMap;  // table - foodMapset<string> foodSet;for (auto &order:orders) {foodSet.insert(order[2]);tableMap[stoi(order[1])][order[2]]++;}vector<vector<string>> result;result.reserve(tableMap.size());vector<string> line;line.reserve(foodSet.size());line.emplace_back("Table");for_each(foodSet.begin(), foodSet.end(), [&](const string &s) {line.emplace_back(s);});result.emplace_back(line);for_each(tableMap.begin(), tableMap.end(), [&](map<int, foodMap>::value_type &f) {line.clear();line.emplace_back(to_string(f.first));for_each(foodSet.begin(), foodSet.end(), [&](const string &s) {line.emplace_back(to_string(f.second[s]));});result.emplace_back(line);});return result;}
};

23-01 字符流

struct Node {bool isEnd;Node *children[26];Node() : isEnd(false) { memset(children, 0, sizeof children); };
};class StreamChecker {private:Node *root;string s;
public:StreamChecker(vector<string> &words) {root = new Node();for (auto &w:words) {auto node = root;for (int i = w.size() - 1; i >= 0; --i) {if (!node->children[w[i] - 'a']) node->children[w[i] - 'a'] = new Node();node = node->children[w[i] - 'a'];}node->isEnd = true;}}bool query(char letter) {s.push_back(letter);auto node = root;for (int i = s.size() - 1; i >= 0; --i) {if (node->isEnd) return true;if (!node->children[s[i] - 'a']) return false;node = node->children[s[i] - 'a'];}return node->isEnd;}
};

23-02 使网格图至少有一条有效路径的最小代价

class Solution {private:int DIR[5][2] = {{0,  0}, {0,  1}, {0,  -1}, {1,  0}, {-1, 0}};public:int minCost(vector<vector<int>> &grid) {int n = grid.size(), m = grid[0].size();vector<vector<int>> cost(n, vector<int>(m, INT_MAX));deque<tuple<int, int, int>> deq;deq.emplace_back(0, 0, 0);while (!deq.empty()) {auto[x, y, c] = deq.front();deq.pop_front();if (c >= cost[x][y]) continue;cost[x][y] = c;for (int i = 1; i <= 4; ++i) {int nx = x + DIR[i][0], ny = y + DIR[i][1];if (nx < 0 || ny < 0 || nx == n || ny == m) continue;if (grid[x][y] == i) {if (cost[nx][ny] > c) deq.emplace_front(nx, ny, c);} else {if (cost[nx][ny] > c + 1) deq.emplace_back(nx, ny, c + 1);}}}return cost[n - 1][m - 1];}
};

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2022-7 刷题记录

20-01. 找到需要补充粉笔的学生编号

20-02. 最小平均差

20-03. 满足三条件之一需改变的最少字符数

20-04. 蜡烛之间的盘子

21-01. 快照数组

21-02. 满足条件的子序列数目

20-03. 有界数组中指定下标处的最大值

21-01. 两个有序数组的第 K 小乘积

21-02. 设计一个文本编辑器

21-03. 翻转等价二叉树

21-04. 找到所有的农场组

21-05. 寻找重复的子树

21-06. 二叉搜索树的范围和

21-07. 所有大于等于节点的值之和

21-08. 把二叉搜索树转换为累加树

21-09. 点菜展示表

23-01 字符流

23-02 使网格图至少有一条有效路径的最小代价

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