$\large\color{orange}{比赛入口：}$$\color{orange}{LeetCode200}$$\large\color{orange}{周赛}$

$\large\color{green}{1534. 统计好三元组}$

枚举，时间复杂度：$O(n^{3})$

class Solution {
public:
    int countGoodTriplets(vector<int>& arr, int a, int b, int c) {
        int n = arr.size();
        int cnt = 0;
        for(int i = 0; i < n; i ++ ){
            for(int j = i + 1; j < n; j ++ ){
                for(int k = j + 1; k < n; k ++ ){
                    cnt += abs(arr[i] - arr[j]) <= a && abs(arr[j] - arr[k]) <= b && abs(arr[i] - arr[k]) <= c ;
                }
            }
        }
        return cnt;
    }
};

$\large\color{orange}{1535. 找出数组游戏的赢家}$

模拟+贪心，时间复杂度：$O(n)$

class Solution {
public:
    unordered_map<int, int> mp;
    int getWinner(vector<int>& arr, int k) {
        int n = arr.size();
        int i = 0, j = 1;
        while(true){
            if(arr[i] > arr[j]){
                mp[arr[i]] ++ ;
                if(mp[arr[i]] == k) return arr[i];
                j ++ ;
            }else{
                mp[arr[j]] ++ ;
                if(mp[arr[j]] == k) return arr[j];
                i = j; j ++ ;
            }
            if(j == n)  return arr[i];
        }
        return 0;
    }
};

$\large\color{orange}{1536. 排布二进制网格的最少交换次数}$

贪心，时间复杂度：$O(n^{2})$

class Solution {
public:
    unordered_map<int, int> mp;
    static const int N = 210;
    int minSwaps(vector<vector<int>>& grid) {
        int n = grid.size();
        for(int i = 0; i < n; i ++ ){
            int cnt = 0;
            bool flag = true;
            for(int j = n - 1; j >= 0; j -- ){
                if(grid[i][j])  flag = false;
                if(flag)    cnt += (grid[i][j] == 0);
            }
            mp[i] = cnt;
        }
        int res = 0;
        int st = 0;
        for(int i = n - 1; i >= 1; i -- ){
            int minv = 0x3f3f3f3f, index = -1;
            for(int j = st; j < n; j ++ ){
                if(mp[j] >= i && minv > j - st) index = j, minv = j - st;  
            }
            if(index == -1) return -1;
            for(int j = index; j > st; j -- )  mp[j] = mp[j - 1];
            res += minv;
            st ++ ;
        }
        return res;

    }
};

$\large\color{red}{1537. 最大得分}$

贪心+双指针，时间复杂度：$O(n+m)$

class Solution {
public:
    typedef long long ll;
    static const ll N = 1E5 + 10, MOD = 1E9 + 7;
    unordered_map<int, int> mp1;
    typedef pair<int, int> pii;
    unordered_map<int, pii> p;
    int maxSum(vector<int>& nums1, vector<int>& nums2) {
        int n1 = nums1.size(), n2 = nums2.size();
        for(int i = 0; i < n1; i ++ )   mp1[nums1[i]] = i;
        for(int i = 0; i < n2; i ++ )   if(mp1.find(nums2[i]) != mp1.end()) p[nums2[i]] = {mp1[nums2[i]], i};
        ll maxv1 = 0;
        ll maxv2 = 0;
        int st0 = 0, st1 = 0;
        while(st0 < n1){
            maxv1 = (maxv1 + nums1[st0]);
            if(p.find(nums1[st0]) != p.end()){
                int ed = p[nums1[st0]].second;
                while(st1 <= ed){
                    maxv2 = (maxv2 + nums2[st1]);
                    st1 ++ ;
                }
                maxv1 = max(maxv1, maxv2);
                maxv2 = maxv1;
            }
            st0 ++ ;
        }
        while(st1 < n2){
            maxv2 = (maxv2 + nums2[st1]);
            st1 ++ ;
        }
        return max(maxv1, maxv2) % MOD;
    }
};