leetcode竞赛：85 场双周赛

链接：https://leetcode.cn/contest/biweekly-contest-85/
日期：2022年08月20日
1.
定长滑动窗口

class Solution {
public:
    int minimumRecolors(string blocks, int k) {
        int n = blocks.size(), cnt = 0, res = 1000;
        for (int i = 0, j = 0; i < n; i++) {
            while (j < n && j - i < k) {
                if (blocks[j++] == 'W') cnt++;
            }
            if (j - i == k) res = min(res, cnt);
            if (blocks[i] == 'W') cnt--;
        }
        return res;
    }
};
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暴力模拟

class Solution {
public:
    int secondsToRemoveOccurrences(string s) {
        int res = 0, n = s.size();
        while (true) {
            int flag = 0;
            for (int i = 1; i < n; i++) {
                if (s[i - 1] == '0' && s[i] == '1') {
                    swap(s[i-1], s[i]);
                    i++;
                    flag = 1;
                }
            }

            if (!flag) break;
            res++;

        }
        return res;
    }
};
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差分算法

class Solution {
public:
    string shiftingLetters(string s, vector<vector<int>>& shifts) {
        int n = s.size();
        vector<int> t(n + 1);
        for (auto &p:shifts) {
            int a = p[0], b = p[1], d = p[2];
            if (d == 0) d = -1;
            t[a] += d;
            t[b + 1] -= d;
        }

        for (int i = 1; i <= n; i++) {
            t[i] += t[i - 1];
            t[i - 1] = t[i - 1] % 26 + 26;
            char c = (s[i - 1] - 'a' + t[i - 1]) % 26 + 'a';
            s[i - 1] = c;
        }
        return s;
    }
};
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逆向并查集 做法

class Solution {
public:
    typedef long long LL;
    vector<LL> s;
    vector<int> p;
    int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
    LL Union(int a, int b) {
        a = find(a), b = find(b);
        p[a] = b;
        s[b] += s[a];
        return s[b];
    }
    vector<long long> maximumSegmentSum(vector<int>& nums, vector<int>& removeQueries) {
        vector<LL> res;
        int n = removeQueries.size();
        for (int i = 0; i < nums.size(); i++) {
            p.push_back(i);
            s.push_back(0);
        }
        LL mx = 0;
        for (int i = n - 1; i >= 0; i--) {
            res.push_back(mx);
            int x = removeQueries[i];
            s[x] = nums[x];
            if (x > 0 && s[x - 1]) mx = max(mx, Union(x, x - 1));
            if (x < n - 1 && s[x + 1]) mx = max(mx, Union(x, x + 1));
            mx = max(mx, s[x]);
        }
        reverse(res.begin(), res.end());
        return res;
    }
};
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模拟列项正向维护分裂数组的最大值：二分，lower_bound

class Solution {
public:
    typedef long long LL;
    vector<long long> maximumSegmentSum(vector<int>& nums, vector<int>& removeQueries) {
        int n = nums.size();
        vector<LL> f(n + 1), res;
        for (int i = 0; i < n; i++) {
            f[i + 1] = f[i] + nums[i];
        }
        set<int> st; 
        st.insert(0); st.insert(n);
        multiset<long long> ms;
        ms.insert(f[n]);
        for (int pos : removeQueries) {
            auto i = st.upper_bound(pos);
            auto j = (i--);
            int L = *i, R = *j;
            ms.erase(ms.lower_bound(f[R] - f[L]));
            st.insert(pos);
            st.insert(pos + 1);
            ms.insert(f[pos] - f[L]);
            ms.insert(f[R] - f[pos + 1]);
            res.push_back(*ms.rbegin());
        }
        
        return res;
    }
};

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