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胶州企业网站设计,首页界面设计,开发平台都有哪些,北京网站建设有哪些lc1198hash统计二维矩阵中所有数字的出现次数#xff0c;找出出现次数等于矩阵行数的最小数字#xff0c;无则返回 -1class Solution { /* 输入#xff1a;mat [[1,2,3,4,5],[2,4,5,8,10],[3,5,7,9,11],[1,3,5,7,9]] 输出#xff1a;5 */ public:int smallestCommonElemen…lc1198hash统计二维矩阵中所有数字的出现次数找出出现次数等于矩阵行数的最小数字无则返回 -1class Solution {/*输入mat [[1,2,3,4,5],[2,4,5,8,10],[3,5,7,9,11],[1,3,5,7,9]]输出5*/public:int smallestCommonElement(vectorvectorint mat){int mmat.size(),nmat[0].size();unordered_mapint,int hash;for(int i0;im;i){for(int j0;jn;j){hash[mat[i][j]];}}int mnINT_MAX;for(auto [a,b]:hash){if(bm) mnmin(mn,a);}return mnINT_MAX?-1:mn;}};lc1918堆tle...二分try答案 越短越合法-滑窗_计数check vs k典题二分猜子数组和的大小滑窗 数出“和不超过这个数”的子数组有多少个不断缩小范围最终找到第k小的和class Solution {typedef long long ll;public:int kthSmallestSubarraySum(vectorint nums, int k){int n nums.size();int l *min_element(nums.begin(), nums.end());int r accumulate(nums.begin(), nums.end(), 0);while (l r) {int m l (r - l) / 2;int c cnt(nums, m);if (c k) l m 1;else r m-1;}return l;}private://越短越合法滑窗int cnt(vectorint nums, int t){int n nums.size(), c 0, s 0, l 0;for (int r 0; r n; r) {s nums[r];while (s t) {s - nums[l];l;}c r - l 1; //不超过的子数组个数}return c;}};lc2743越短越合法 滑窗ret (r - l 1);class Solution {public:int numberOfSpecialSubstrings(string s) {int n s.size();int l 0, ret 0;unordered_mapchar, int hash;for (int r 0; r n; r) {while (hash.count(s[r]) hash[s[r]] 1) {hash[s[l]]--;l;}hash[s[r]] 1;ret (r - l 1);//cal}return ret;}};lc3652前缀和_分为了3部分预处理两个前缀和(ps_sum,​​​​ p_sum)枚举修改的子数组改后利润改前两边利润(ps_sum)改后子数组的部分售价和(按照题目规则可知: 为k/2后半部分p_sum)取最大利润。class Solution {public:long long maxProfit(vectorint prices, vectorint strategy, int k){int n prices.size();vectorlong long sum(n 1), sum_sell(n 1);for (int i 0; i n; i) {sum[i 1] sum[i] prices[i] * strategy[i];sum_sell[i 1] sum_sell[i] prices[i];}long long ans sum[n]; // 不修改for (int i k; i n; i) {long long res sum[i - k] sum[n] - sum[i] sum_sell[i] - sum_sell[i - k / 2];ans max(ans, res);}return ans;}};