A 找出满足差值条件的下标 I

class Solution {
public:
vector<int> findIndices(vector<int> &nums, int indexDifference, int valueDifference) {
int n = nums.size();
for (int i = 0; i < n; i++)
for (int j = 0; j <= i; j++)
if (i - j >= indexDifference && abs(nums[i] - nums[j]) >= valueDifference)
return {i, j};
return {-1, -1};
}
};
B 最短且字典序最小的美丽子字符串

class Solution {
public:
string shortestBeautifulSubstring(string s, int k) {
int n = s.size();
vector<int> ps(n + 1);
for (int i = 0; i < n; i++)
ps[i + 1] = ps[i] + (s[i] == '1' ? 1 : 0);
for (int len = k; len <= n; len++) {
int find = 0;
for (int i = 0, j = i + len - 1; j < n; i++, j++) {
if (ps[j + 1] - ps[i] == k)
find = 1;
}
string res = "";
if (find) {
for (int i = 0, j = i + len - 1; j < n; i++, j++) {
if (ps[j + 1] - ps[i] == k) {
if (res.empty() || s.substr(i, len) < res)
res = s.substr(i, len);
}
}
return res;
}
}
return "";
}
};
C 找出满足差值条件的下标 II

class Solution {
public:
vector<int> findIndices(vector<int> &nums, int indexDifference, int valueDifference) {
int n = nums.size();
int mx[n], mn[n];
mx[0] = 0;
mn[0] = 0;
for (int i = 1; i < n; i++) {
mx[i] = nums[mx[i - 1]] > nums[i] ? mx[i - 1] : i;
mn[i] = nums[mn[i - 1]] < nums[i] ? mn[i - 1] : i;
}
for (int i = 0; i < n; i++) {
if (i - indexDifference >= 0) {
if (abs(nums[i] - nums[mx[i - indexDifference]]) >= valueDifference)
return {i, mx[i - indexDifference]};
if (abs(nums[i] - nums[mn[i - indexDifference]]) >= valueDifference)
return {i, mn[i - indexDifference]};
}
}
return {-1, -1};
}
};
D 构造乘积矩阵

class Solution {
public:
using ll = long long;
vector<vector<int>> constructProductMatrix(vector<vector<int>> &grid) {
int m = grid.size(), n = grid[0].size();
int mod = 12345;
vector<vector<int>> res(m, vector<int>(n));
vector<ll> row(m, 1), pre(m, 1), suf(m, 1);
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++)
row[i] = row[i] * grid[i][j] % mod;
pre[i] = i != 0 ? pre[i - 1] * row[i] % mod : row[i];
}
for (int i = m - 1; i >= 0; i--)
suf[i] = i != m - 1 ? suf[i + 1] * row[i] % mod : row[i];
vector<ll> left(n), right(n);
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++)
left[j] = j != 0 ? left[j - 1] * grid[i][j] % mod : grid[i][j];
for (int j = n - 1; j >= 0; j--)
right[j] = j != n - 1 ? right[j + 1] * grid[i][j] % mod : grid[i][j];
int other = (i != 0 ? pre[i - 1] : 1) * (i != m - 1 ? suf[i + 1] : 1) % mod;
for (int j = 0; j < n; j++)
res[i][j] = other * (j != 0 ? left[j - 1] : 1) % mod * (j != n - 1 ? right[j + 1] : 1) % mod;
}
return res;
}
};