A 统计特殊字母的数量 I
哈希：遍历然后枚举
cpp
class Solution {
public:
int numberOfSpecialChars(string word) {
unordered_map<char, int> m;
for (auto ch : word)
m[ch] = 1;
int res = 0;
for (char ch = 'a'; ch <= 'z'; ch++)
if (m.count(ch) && m.count('A' + ch - 'a'))
res++;
return res;
}
};B count-the-number-of-special-characters-ii/
哈希：遍历记录各小写字母的最后出现下标，及各大写字母的第一次出现的下标，然后枚举
cpp
class Solution {
public:
int numberOfSpecialChars(string word) {
unordered_map<char, int> m;
for (int i = 0; i < word.size(); i++)
if (word[i] >= 'a' && word[i] <= 'z' || m.find(word[i]) == m.end())
m[word[i]] = i;
int res = 0;
for (char ch = 'a'; ch <= 'z'; ch++)
if (m.count(ch) && m.count('A' + ch - 'a') && m[ch] < m['A' + ch - 'a'])
res++;
return res;
}
};C 使矩阵满足条件的最少操作次数
动态规划：设 p [ i ] [ j ] p[i][j] p[i][j] 为使 g r i d grid grid 的前 i + 1 i+1 i+1 列行成的子矩阵满足条件的且最后一列都为 j j j 的最少操作数，最终答案为 m i n { p [ n − 1 ] [ j ] ∣ 0 ≤ j ≤ 9 } min\{ p[n-1][j] \;|\; 0\le j\le 9 \} min{p[n−1][j]∣0≤j≤9}
cpp
class Solution {
public:
int minimumOperations(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
vector<vector<int>> col(n, vector<int>(10));//col[i][j]: 第i列中j的数目
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
col[j][grid[i][j]]++;
vector<vector<int>> p(n, vector<int>(10));
for (int j = 0; j < 10; j++)
p[0][j] = m - col[0][j];
for (int j = 1; j < n; j++) {
for (int cur = 0; cur < 10; cur++) {
p[j][cur] = INT32_MAX;
for (int last = 0; last < 10; last++)//枚举前一行
if (last != cur)
p[j][cur] = min(p[j][cur], p[j - 1][last] + m - col[j][cur]);
}
}
int res = INT32_MAX;
for (int v = 0; v < 10; v++)
res = min(res, p[n - 1][v]);
return res;
}
};D 最短路径中的边
最短路 + b f s bfs bfs：设 d [ i ] d[i] d[i] 为 0 0 0 到 i i i 的最短路的长度，先用最 d i j k s t r a dijkstra dijkstra 求 0 0 0 到 n − 1 n-1 n−1 的最短路。如果 0 0 0 与 n − 1 n-1 n−1 连通，则从 n − 1 n-1 n−1 开始 b f s bfs bfs：设 b f s bfs bfs 当前遍历到的节点为 i i i，若 i i i 的邻接点 j j j 满足 d [ j ] + w { j , i } = d [ i ] d[j]+w_{\{j,i\}}=d[i] d[j]+w{j,i}=d[i] ，则将 a n s w e r answer answer 数组中 ( j , i ) (j,i) (j,i) 边对应的下标位置置为 t r u e true true ，同时将 j j j 加入 b f s bfs bfs 队列。
cpp
class Solution {
public:
using ll = long long;
vector<bool> findAnswer(int n, vector<vector<int>>& edges) {
vector<vector<pair<int, int>>> e(n);
map<pair<int, int>, int> id;//记录各表在edges中的下标
for (int i = 0; i < edges.size(); i++) {//建图
auto& edge = edges[i];
e[edge[0]].push_back({edge[1], edge[2]});
e[edge[1]].push_back({edge[0], edge[2]});
id[{min(edge[0], edge[1]), max(edge[0], edge[1])}] = i;
}
vector<ll> d(n, INT64_MAX);
d[0] = 0;
priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> q;//最小堆
q.emplace(0, 0);
while (!q.empty()) {//求最短路
auto [di, i] = q.top();
q.pop();
for (auto [j, w] : e[i]) {
if (d[j] > di + w) {
d[j] = di + w;
q.emplace(d[j], j);
}
}
}
vector<int> vis(n);
queue<int> qu;
vector<bool> res(edges.size());
if (d[n - 1] != INT64_MAX)
qu.push(n - 1);
while (!qu.empty()) {//bfs
auto i = qu.front();
qu.pop();
for (auto [j, w] : e[i]) {
if (d[j] != INT64_MAX && d[j] + w == d[i]) {
res[id[make_pair(min(i, j), max(i, j))]] = true;
if (!vis[j]) {
vis[j] = 1;//入队标记
qu.push(j);
}
}
}
}
return res;
}
};