力扣第374场周赛

找出峰值

模拟

class Solution {
public:vector<int> findPeaks(vector<int>& mountain) {vector<int>ans;for(int i = 1 ; i < mountain.size() - 1; i ++){if(mountain[i] > mountain[i-1] && mountain[i] > mountain[i+1]){ans.push_back(i);}}return ans;}
};

需要添加的硬币的最小数量

贪心遍历，假设现在得到了0~(s-1)内的所有整数，如果此时新发现了一个整数 x，那么把x加到已得到的数字中，就得到了x~(s+x−1)内的所有整数。

class Solution {
public:int minimumAddedCoins(vector<int>& coins, int target) {sort(coins.begin() , coins.end());//当前0~cur的值都可以得到sort(coins.begin(), coins.end());long long ans = 0 , cur = 0 , n = coins.size();for (int x : coins) {while (x > cur + 1) {cur += cur + 1;ans++;}cur += x;}while (cur < target) {cur += cur + 1;ans++;}return ans;}
};

统计完全子字符串

条件二分段，条件一滑窗统计

class Solution {int f(string s, int k) {int res = 0;for (int m = 1; m <= 26 && k * m <= s.length(); m++) {int cnt[26]{};auto check = [&]() {for (int i = 0; i < 26; i++) {if (cnt[i] && cnt[i] != k) {return;}}res++;};for (int right = 0; right < s.length(); right++) {// 滑窗记录个数cnt[s[right] - 'a']++;int left = right + 1 - k * m;if (left >= 0) {check();//是否满足条件一cnt[s[left] - 'a']--;}}}return res;}public:int countCompleteSubstrings(string word, int k) {int n = word.length();int ans = 0;for (int i = 0; i < n;) {int st = i;//分段for (i++; i < n && abs(int(word[i]) - int(word[i - 1])) <= 2; i++);ans += f(word.substr(st, i - st), k);}return ans;}
};

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