Commit 3743a6c

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add 102, 965, 1876

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Lines changed: 41 additions & 0 deletions

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	`1`	`+/*`
	`2`	`+102. Binary Tree Level Order Traversal`
	`3`	`+`
	`4`	`+Submitted: January 13, 2024`
	`5`	`+`
	`6`	`+Runtime: 0 ms (beats 100.00%)`
	`7`	`+Memory: 17.24 MB (beats 14.20%)`
	`8`	`+*/`
	`9`	`+`
	`10`	`+/**`
	`11`	`+ * Definition for a binary tree node.`
	`12`	`+ * struct TreeNode {`
	`13`	`+ * int val;`
	`14`	`+ * TreeNode *left;`
	`15`	`+ * TreeNode *right;`
	`16`	`+ * TreeNode() : val(0), left(nullptr), right(nullptr) {}`
	`17`	`+ * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}`
	`18`	`+ * TreeNode(int x, TreeNode left, TreeNode right) : val(x), left(left), right(right) {}`
	`19`	`+ * };`
	`20`	`+ */`
	`21`	`+class Solution {`
	`22`	`+public:`
	`23`	`+ vector<vector<int>> levelOrder(TreeNode* root) {`
	`24`	`+ if (root == nullptr) return {};`
	`25`	`+ vector<vector<int>> res;`
	`26`	`+ deque<TreeNode*> queue;`
	`27`	`+ queue.push_front(root);`
	`28`	`+ while (!queue.empty()) {`
	`29`	`+ vector<int> v;`
	`30`	`+ for (int i = 0, n = queue.size(); i < n; ++i) {`
	`31`	`+ TreeNode* p = queue.back();`
	`32`	`+ queue.pop_back();`
	`33`	`+ v.push_back(p->val);`
	`34`	`+ if (p->left != nullptr) queue.push_front(p->left);`
	`35`	`+ if (p->right != nullptr) queue.push_front(p->right);`
	`36`	`+ }`
	`37`	`+ res.push_back(v);`
	`38`	`+ }`
	`39`	`+ return res;`
	`40`	`+ }`
	`41`	`+};`

Lines changed: 25 additions & 0 deletions

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	`1`	`+/*`
	`2`	`+1876. Substrings of Size Three with Distinct Characters`
	`3`	`+`
	`4`	`+Submitted: January 13, 2024`
	`5`	`+`
	`6`	`+Runtime: 0 ms (beats 100.00%)`
	`7`	`+Memory: 8.37 MB (beats 73.21%)`
	`8`	`+*/`
	`9`	`+`
	`10`	`+class Solution {`
	`11`	`+public:`
	`12`	`+ int countGoodSubstrings(string s) {`
	`13`	`+ if (s.size() < 3) return 0;`
	`14`	`+ // char arr[3];`
	`15`	`+ int res = 0;`
	`16`	`+ for (int i = 0, n = s.size() - 2; i < n; ++i) {`
	`17`	`+ res += (`
	`18`	`+ s[i] != s[i + 1] && s[i] != s[i + 2] &&`
	`19`	`+ s[i + 1] != s[i] && s[i + 1] != s[i + 2] &&`
	`20`	`+ s[i + 2] != s[i] && s[i + 2] != s[i + 1]`
	`21`	`+ );`
	`22`	`+ }`
	`23`	`+ return res;`
	`24`	`+ }`
	`25`	`+};`

Lines changed: 12 additions & 0 deletions

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	`1`	`+"""`
	`2`	`+1876. Substrings of Size Three with Distinct Characters`
	`3`	`+`
	`4`	`+Submitted: January 13, 2024`
	`5`	`+`
	`6`	`+Runtime: 0 ms (beats 100.00%)`
	`7`	`+Memory: 17.72 MB (beats 21.15%)`
	`8`	`+"""`
	`9`	`+`
	`10`	`+class Solution:`
	`11`	`+ def countGoodSubstrings(self, s: str) -> int:`
	`12`	`+ return sum(len(set(s[x:x+3])) == 3 for x in range(0, len(s) - 2))`

Lines changed: 31 additions & 0 deletions

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	`1`	`+/*`
	`2`	`+965. Univalued Binary Tree`
	`3`	`+`
	`4`	`+Submitted: January 13, 2024`
	`5`	`+`
	`6`	`+Runtime: 0 ms (beats 100.00%)`
	`7`	`+Memory: 12.69 MB (beats 29.98%)`
	`8`	`+*/`
	`9`	`+`
	`10`	`+/**`
	`11`	`+ * Definition for a binary tree node.`
	`12`	`+ * struct TreeNode {`
	`13`	`+ * int val;`
	`14`	`+ * TreeNode *left;`
	`15`	`+ * TreeNode *right;`
	`16`	`+ * TreeNode() : val(0), left(nullptr), right(nullptr) {}`
	`17`	`+ * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}`
	`18`	`+ * TreeNode(int x, TreeNode left, TreeNode right) : val(x), left(left), right(right) {}`
	`19`	`+ * };`
	`20`	`+ */`
	`21`	`+`
	`22`	`+class Solution {`
	`23`	`+public:`
	`24`	`+ int val = -1;`
	`25`	`+ bool isUnivalTree(TreeNode* root) {`
	`26`	`+ if (root == nullptr) return true;`
	`27`	`+ if (val == -1) val = root->val;`
	`28`	`+ if (root->val != val) return false;`
	`29`	`+ return isUnivalTree(root->left) && isUnivalTree(root->right);`
	`30`	`+ }`
	`31`	`+};`

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