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A 最长的严格递增或递减子数组
模拟:实现可以只求最长严格递增子数组即可
class Solution {public:int longestIncrementsubarray(vector<int> &nums) {int n = nums.size();int res = 1;for (int i = 0, j = 0; i < n; i = ++j) {while (j + 1 < n && nums[j + 1] > nums[j])j++;res = max(res, j - i + 1);}return res;}int longestMonotonicSubarray(vector<int> &nums) {int res = longestIncrementsubarray(nums);reverse(nums.begin(), nums.end());res = max(res, longestIncrementsubarray(nums));return res;}
};
B 满足距离约束且字典序最小的字符串
贪心:尽可能将先前面的字符变小
class Solution {public:string getSmallestString(string s, int k) {for (int i = 0; i < s.size() && k; i++) {if (s[i] != 'a') {//s[i]可以变小int d = min(k, min(s[i] - 'a', 26 - (s[i] - 'a')));//s[i]的操作数k -= d;s[i] = min(s[i] - d, 'a' + (s[i] - 'a' + d) % 26);}}return s;}
};
C 使数组中位数等于 K 的最少操作数
排序+模拟:先将数组排序,设中位数的下标为 m i d mid mid , n u m s [ m i d ] nums[mid] nums[mid] 需要 ∣ k − n u m s [ m i d ] ∣ |k-nums[mid]| ∣k−nums[mid]∣ 次操作变为 k k k 。顺序枚举 n u m s [ m i d + 1 , n u m . s i z e ( ) − 1 ] nums[mid+1,num.size()-1] nums[mid+1,num.size()−1] ,若 n u m s [ i ] nums[i] nums[i] 比 n u m s [ i − 1 ] nums[i-1] nums[i−1] 小,则将 n u m s [ i ] nums[i] nums[i] 变为 n u m s [ i − 1 ] nums[i-1] nums[i−1] 。逆序枚举 n u m s [ 0 , m i d − 1 ] nums[0,mid-1] nums[0,mid−1] ,若 n u m s [ i ] nums[i] nums[i] 比 n u m s [ i + 1 ] nums[i+1] nums[i+1] 大,则将 n u m s [ i ] nums[i] nums[i] 变为 n u m s [ i + 1 ] nums[i+1] nums[i+1]。
class Solution {public:long long minOperationsToMakeMedianK(vector<int> &nums, int k) {sort(nums.begin(), nums.end());int n = nums.size();int mid = n / 2;long long res = abs(nums[mid] - k);nums[mid] = k;for (int i = mid + 1; i < n; i++) {if (nums[i] < nums[i - 1]) {res += nums[i - 1] - nums[i];nums[i] = nums[i - 1];}}for (int i = mid - 1; i >= 0; --i) {if (nums[i] > nums[i + 1]) {res += nums[i] - nums[i + 1];nums[i] = nums[i + 1];}}return res;}
};
D 带权图里旅途的最小代价
脑筋急转弯+并查集:先用并查集维护哪些节点在同一个连通分支,在同一分支的 s i s_i si 和 t i t_i ti 的最小代价即为所在连通分支上所有边的按位与。
class Solution {public:vector<int> p;vector<int> cost;int find(int x) {//查找x所在分支的根节点if (p[x] == x)return x;return p[x] = find(p[x]);}void comb(int x, int y) {//合并x和y所在的分支int xx = find(x);int yy = find(y);if (xx != yy)p[yy] = xx;}vector<int> minimumCost(int n, vector<vector<int>> &edges, vector<vector<int>> &query) {p.resize(n);iota(p.begin(), p.end(), 0);//初始化:每个节点为一个分支for (auto &edge : edges)comb(edge[0], edge[1]);cost = vector<int>(n, (1 << 30) - 1);for (auto &edge : edges) {int root = find(edge[0]);cost[root] = cost[root] & edge[2];}vector<int> res;for (auto &q : query) {int r1 = find(q[0]), r2 = find(q[1]);if (r1 != r2)res.push_back(-1);elseres.push_back(cost[r1]);}return res;}
};
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