可以做免费广告的网站运营商大数据精准营销

第 119 场双周赛

找到两个数组中的公共元素

模拟

class Solution {
public:vector<int> findIntersectionValues(vector<int>& nums1, vector<int>& nums2) {unordered_map<int,int> mp1 , mp2;vector<int>res(2);for(auto x : nums1){mp1[x] ++;}for(auto x : nums2){mp2[x] ++;}int k = 0;for(auto x : nums1){if(mp2[x] > 0)k ++;}res[0] = k;k = 0;for(auto x : nums2){if(mp1[x] > 0)k ++;}res[1] = k;return res;}
};

消除相邻近似相等字符

贪心，一个变了下一个肯定不用变

class Solution {
public:int removeAlmostEqualCharacters(string word) {int n = word.size() , ans = 0;for(int i = 1 ; i < n ; i ++)if(abs(word[i] - word[i-1]) <= 1)ans ++,i ++;return ans;}
};

最多 K 个重复元素的最长子数组

滑动窗扣维持k个

class Solution {
public:int maxSubarrayLength(vector<int>& nums, int k) {int n = nums.size();unordered_map<int,int>m;queue<int>q;int ans = 0;for(int i = 0 ; i < n ; i ++){m[nums[i]] ++;q.push(nums[i]);while(m[nums[i]] > k){int t = q.front();q.pop();m[t] --;}int l = q.size();ans = max(ans , l);}return ans;}
};

找到最大非递减数组的长度

二进制枚举+Floyd

class Solution {
public:static const int N = 11;int d[N][N];void init(){for(int i = 0 ; i < N ; i ++){for(int j = 0 ; j < N ; j ++){if(i == j)d[i][j] = 0;else d[i][j] = 99999;}}}int numberOfSets(int n, int md, vector<vector<int>>& roads) {int ans = 1 , cnt = roads.size();for(int i = 1 ; i <= ((1 << n) - 1) ; i ++){ //枚举//建图init();vector<int> c;vector<int> nt(15);for(int j = 0 ; j <= 10 ; j ++){if(i >> j & 1)c.push_back(j);}for(auto x : c)nt[x] = 1;for(int j = 0 ; j < cnt ; j ++){if(nt[roads[j][0]] == 1 && nt[roads[j][1]] == 1){d[roads[j][0]][roads[j][1]] = min(d[roads[j][0]][roads[j][1]],roads[j][2]);d[roads[j][1]][roads[j][0]] = min(d[roads[j][1]][roads[j][0]],roads[j][2]);                    }}function<bool()> check = [&]() -> bool{for (int kk = 0; kk < n; kk ++ )for (int ii = 0; ii < n; ii ++ )for (int jj = 0; jj < n; jj ++ )d[ii][jj] = min(d[ii][jj], d[ii][kk] + d[kk][jj]);for(int a = 0 ; a < c.size() ; a ++){for(int b = 0 ; b < c.size() ; b ++){if(d[c[a]][c[b]] > md){return false;}}}return true;};if(check())ans ++;} return ans;}
};

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