你可不可以
成为我的main函数
做我此生必须有
且只能有一个的入口
——————————
我愿为自己加上private
在你的class中只有
你能调用
——————————————代 码 如 诗 。
Given an integer (signed 32 bits), write a function to check whether it is a power of 4.
Example:
Given num = 16, return true. Given num = 5, return false.
Follow up: Could you solve it without loops/recursion?
Credits:
Special thanks to @yukuairoy for adding this problem and creating all test cases.
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class Solution { public: bool isPowerOfFour(int num) { return num <= 0 ? false : pow(4, (round)(log(num) / log(4))) == num; } }; |
Write a function to find the longest common prefix string amongst an array of strings.
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class Solution { public: string longestCommonPrefix(vector<string>& strs) { if(strs.size() == 0) return ""; string s = strs[0]; for(int i = 1; i < strs.size(); i++) { for(int j = 0; j < s.length(); j++) { if(s[j] != strs[i][j]) { s = s.substr(0, j); break; } } } return s; } }; |
Remove all elements from a linked list of integers that have value val.
Example
Given: 1 –> 2 –> 6 –> 3 –> 4 –> 5 –> 6, val = 6
Return: 1 –> 2 –> 3 –> 4 –> 5
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/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* removeElements(ListNode* head, int val) { while(head != NULL && head->val == val) { head = head->next; } if(head == NULL) { return NULL; } ListNode *p = head; ListNode *q= p->next; while(p != NULL && p->next != NULL) { q = p->next; while(q != NULL && q->val == val) { q = q->next; } p->next = q; p = q; } return head; } }; |
203. Remove Linked List Elements
My Submissions QuestionEditorial Solution
Total Accepted: 59711 Total Submissions: 209326 Difficulty: Easy
Remove all elements from a linked list of integers that have value val.
Example
Given: 1 –> 2 –> 6 –> 3 –> 4 –> 5 –> 6, val = 6
Return: 1 –> 2 –> 3 –> 4 –> 5
Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.
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/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* removeElements(ListNode* head, int val) { while(head != NULL && head->val == val) { head = head->next; } if(head == NULL) { return NULL; } ListNode *p = head; ListNode *q= p->next; while(p != NULL && p->next != NULL) { q = p->next; while(q != NULL && q->val == val) { q = q->next; } p->next = q; p = q; } return head; } }; |
The count-and-say sequence is the sequence of integers beginning as follows:
1, 11, 21, 1211, 111221, …
1 is read off as “one 1” or 11.
11 is read off as “two 1s” or 21.
21 is read off as “one 2, then one 1” or 1211.
Given an integer n, generate the nth sequence.
Note: The sequence of integers will be represented as a string.
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class Solution { public: string countAndSay(int n) { string temp = ""; string num = ""; int cnt = 1; if(n == 1) { return "1"; } if(n == 2) { return "11"; } string a = "11"; for(int i = 1; i <= n - 2; i++) { for(int j = 1; j < a.length(); j++) { if(a[j-1] == a[j]) { cnt++; } if(a[j-1] != a[j]) { while(cnt) { num = (char)(cnt % 10 + '0') + num; cnt = cnt / 10; } temp += num; num = ""; temp += a[j-1]; cnt = 1; } if(j == a.length() - 1) { while(cnt) { num = (char)(cnt % 10 + '0') + num; cnt = cnt / 10; } temp += num; num = ""; temp += a[j]; cnt = 1; } } a = temp; temp = ""; } return a; } }; |
Given a pattern and a string str, find if str follows the same pattern.
Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in str.
Examples:
pattern = “abba”, str = “dog cat cat dog” should return true.
pattern = “abba”, str = “dog cat cat fish” should return false.
pattern = “aaaa”, str = “dog cat cat dog” should return false.
pattern = “abba”, str = “dog dog dog dog” should return false.
Notes:
You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.
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class Solution { public: bool wordPattern(string pattern, string str) { map<char, string> ptos; map<string, char> stop; int len = pattern.length(); string *s = new string [len]; int cnt = 0; for(int i = 0; i < str.length(); i++) { if(str[i] == ' ') cnt++; } if(cnt != len - 1) { return false; } int t = 0; for(int i = 0; i < str.length(); i++) { if(str[i] != ' ') { s[t] += str[i]; } else { t++; } } for(int i = 0; i < len; i++) { if(ptos.find(pattern[i]) != ptos.end() && ptos[pattern[i]] != s[i] || stop.find(s[i]) != stop.end() && stop[s[i]] != pattern[i]) return false; ptos[pattern[i]] = s[i]; stop[s[i]] = pattern[i]; } return true; } }; |