LeetCode 第 212 场周赛
作者:互联网
枚举模拟
1 class Solution { 2 public: 3 char slowestKey(vector<int>& releaseTimes, string keysPressed) { 4 map<char, int> m; 5 int mx = 0; 6 for (int i = 0; i < releaseTimes.size(); i++) { 7 if (i == 0) { 8 m[keysPressed[i]] = releaseTimes[i]; 9 } else { 10 m[keysPressed[i]] = max(m[keysPressed[i]], releaseTimes[i] - releaseTimes[i - 1]); 11 } 12 mx = max(mx, m[keysPressed[i]]); 13 } 14 char ans = 'a'; 15 for (int i = 0; i < keysPressed.size(); i++) { 16 if (m[keysPressed[i]] == mx) { 17 ans = max(ans, keysPressed[i]); 18 } 19 } 20 return ans; 21 } 22 };View Code
B 等差子数组
枚举模拟 时间复杂度O(n^2 * logn)
1 class Solution { 2 public: 3 vector<bool> checkArithmeticSubarrays(vector<int>& nums, vector<int>& l, vector<int>& r) { 4 vector<int> v; 5 vector<bool> ans; 6 for (int i = 0; i < l.size(); i++) { 7 v.clear(); 8 for (int j = l[i]; j <= r[i]; j++) v.push_back(nums[j]); 9 sort(v.begin(), v.end()); 10 for (int j = 2; j < v.size(); j++) { 11 if ((v[j] - v[j - 1]) != (v[1] - v[0])) { 12 ans.push_back(false); 13 break; 14 } 15 } 16 if (ans.size() != i + 1) ans.push_back(true); 17 } 18 return ans; 19 } 20 };View Code
标签:周赛,releaseTimes,212,int,mx,vector,keysPressed,ans,LeetCode 来源: https://www.cnblogs.com/pavtlly/p/13972147.html