力扣374周赛

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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