暴力枚举

class Solution {
public:
    int alternatingSubarray(vector<int> &nums) {
        int n = nums.size();
        for (int len = n; len > 1; len--) {
            for (int i = 0; i + len - 1 < n; i++) {
                int j = i + 1;
                for (; j < i + len; j++)
                    if (nums[j] - nums[j - 1] != ((j - i) % 2 ? 1 : -1))
                        break;
                if (j == i + len)
                    return len;
            }
        }
        return -1;
    }
};

有序集合模拟

class Solution {
public:
    vector<int> relocateMarbles(vector<int> &nums, vector<int> &moveFrom, vector<int> &moveTo) {
        set<int> s{nums.begin(), nums.end()};
        for (int i = 0; i < moveFrom.size(); i++) {
            s.erase(moveFrom[i]);
            s.insert(moveTo[i]);
        }
        return vector<int>{s.begin(), s.end()};
    }
};

动态规划: 设$s$长为$n$, 定义$p_l$为将$s[l,n)$分割为最少的美丽子字符串的子串数, 若$s[l,n)$对应的数为5的幂则$p_l$=1, 否则有转移方程: p l = m i n { p r + 1    ∣    l < r < n    且    s [ l , r ) 对应的数是 5 的幂 } p_l= min\{p_{r}+1 \;|\; l<r<n \; 且 \; s[l,r)对应的数是5的幂 \} pl​=min{pr​+1∣l<r<n且s[l,r)对应的数是5的幂}

class Solution {
public:
    int getval(string s) {//2进制串转10进制数
        int cur = 0;
        for (auto c: s)
            cur = cur * 2 + c - '0';
        return cur;
    }

    bool ispow5(int val) {//判断是否为5的非负整数幂
        for (; val % 5 == 0; val /= 5);
        return val == 1;
    }

    int minimumBeautifulSubstrings(string s) {
        int n = s.size();
        vector<int> p(n, INT32_MAX);//初始化状态为无效
        if (s[n - 1] == '1')
            p[n - 1] = 1;
        for (int l = n - 2; l >= 0; l--)
            if (s[l] != '0') {//不能有前导零
                if (ispow5(getval(s.substr(l))))
                    p[l] = 1;
                for (int r = l + 1; r < n; r++)
                    if (p[r] != INT32_MAX) {//p[r]状态有效
                        if (ispow5(getval(s.substr(l, r - l))))
                            p[l] = min(p[l], p[r] + 1);
                    }
            }
        return p[0] == INT32_MAX ? -1 : p[0];
    }
};

哈希: 枚举coordinates中的黑色格子, 每个黑色格子最多能使周围的4个块的块内黑格子数加1, 用哈希记录出现过的块的块内黑格子数.

class Solution {
public:
    vector<long long> countBlackBlocks(int m, int n, vector<vector<int>> &coordinates) {
        map<pair<int, int>, int> cnt;//块左上角的坐标 -> 块内黑格子数
        int dx[4] = {0, -1, -1, 0};
        int dy[4] = {-1, -1, 0, 0};
        for (auto &p: coordinates) {
            for (int k = 0; k < 4; k++) {
                int nx = p[0] + dx[k];
                int ny = p[1] + dy[k];
                if (nx < 0 || nx >= m - 1 || ny < 0 || ny >= n - 1)
                    continue;
                cnt[{nx, ny}]++;//块左上角的坐标为(nx,ny)的块的块内黑格子数更新
            }
        }
        vector<long long> res(5);
        for (auto &[_, v]: cnt)
            res[v]++;
        res[0] = 1LL * (m - 1) * (n - 1) - accumulate(res.begin() + 1, res.end(), 0);
        return res;
    }
};