模拟

class Solution {
public:vector<int> resultArray(vector<int> &nums) {vector<int> r1{nums[0]}, r2{nums[1]};for (int i = 2; i < nums.size(); i++) {if (r1.back() > r2.back())r1.push_back(nums[i]);elser2.push_back(nums[i]);}for (auto x: r2)r1.push_back(x);return r1;}
};

前缀和：先计算二维前缀和，再枚举包含左上角元素的子矩阵

class Solution {
public:int countSubmatrices(vector<vector<int>> &grid, int k) {int m = grid.size(), n = grid[0].size();for (int i = 0; i < m; i++)for (int j = 1; j < n; j++)grid[i][j] += grid[i][j - 1];for (int j = 0; j < n; j++)for (int i = 1; i < m; i++)grid[i][j] += grid[i - 1][j];int res = 0;for (int i = 0; i < m; i++)for (int j = 0; j < n; j++)if (grid[i][j] <= k)res++;return res;}
};

枚举：枚举属于 Y 和不属于 Y 的单元格的颜色

class Solution {
public:int minimumOperationsToWriteY(vector<vector<int>> &grid) {int n = grid.size();int res = INT32_MAX;for (int cy = 0; cy <= 2; cy++)//属于Y的单元格的颜色for (int cny = 0; cny <= 2; cny++)//不属于Y的单元格的颜色if (cy != cny) {int t = 0;for (int i = 0; i < n; i++)for (int j = 0; j < n; j++)if (i <= n / 2 && (i == j || i + j == n - 1) || i > n / 2 && j == n / 2)//属于Y的单元格t += grid[i][j] != cy ? 1 : 0;elset += grid[i][j] != cny ? 1 : 0;res = min(res, t);}return res;}
};

离散化 + 树状数组：先将 nums 离散化，然后利用树状数组来维护两个数组，并查询数组中严格大于某个数的元素数量

class Solution {
public:vector<int> resultArray(vector<int> &nums) {vector<int> li = nums, o = li;sort(o.begin(), o.end());o.erase(unique(o.begin(), o.end()), o.end());for (auto &i: li)//li为nums离散化后的数组i = lower_bound(o.begin(), o.end(), i) - o.begin() + 1;int n = nums.size();int m = li.size();vector<int> r1{nums[0]}, r2{nums[1]};BinaryIndexedTree t1(m), t2(m);//两个树状数组t1.add(li[0], 1);t2.add(li[1], 1);for (int i = 2; i < n; i++) {int c1 = t1.query(m) - t1.query(li[i]);int c2 = t2.query(m) - t2.query(li[i]);if (c1 > c2) {r1.push_back(nums[i]);t1.add(li[i], 1);} else if (c1 < c2) {r2.push_back(nums[i]);t2.add(li[i], 1);} else {if (r1.size() <= r2.size()) {r1.push_back(nums[i]);t1.add(li[i], 1);} else {r2.push_back(nums[i]);t2.add(li[i], 1);}}}for (auto x: r2)r1.push_back(x);return r1;}class BinaryIndexedTree {//树状数组模板public:int N;vector<int> a;BinaryIndexedTree(int n) {N = n;a = vector<int>(N + 1);}inline int lowbit(int x) {return x & -x;}void add(int loc, int val) {// li[loc]+=val;for (; loc <= N; loc += lowbit(loc))a[loc] += val;}int query(int loc) {// sum{li[k] | 1<=k<=loc}int res = 0;for (; loc > 0; loc -= lowbit(loc))res += a[loc];return res;}};
};