pq|消消乐|定长滑窗

lc2067

固定长度滑动窗口，遍历所有“k种字符各出现count次”的子串长度（k*count）

统计符合条件的子串总数

优雅变量的控制 只能不断的练习与感受...

class Solution {
public:
int equalCountSubstrings(string s, int cnt)

{
int n = s.size(), res = 0;
for (int k = 1; k <= 26; k++) {
int len = k * cnt;
if (len > n) break;

vector<int> c(26, 0);
int v = 0;
for (int i = 0; i < len; i++)
if (++c[s[i]-'a'] == cnt) v++;
if (v == k) res++;

for (int i = len; i < n; i++) {
int out = s[i-len]-'a', in = s[i]-'a';
if (c[out] == cnt) v--;
c[out]--;
if (++c[in] == cnt) v++;
if (v == k) res++;
}
}
return res;
}
};

lc723

“糖果消消乐”的实现：

**标记负数（替代额外数组）**找出行/列中连续3个相同的糖果

让未消除的糖果下落补位，重复直到没有可消除的糖果

1.总体思路：就是消消乐

• 模拟过程
• 标记
• 整理

2.利用数值的正负性，来省略调mark数组

3.整理的时候注意从下往上收。

桶排序（bucketSort）也是类似的整理顺序

class Solution
{
public:
vector<vector<int>> candyCrush(vector<vector<int>>& board)
{
int Row = board.size(), Col = board[0].size();
bool need_todo = true;
//////// 思路：根据例子，L形也是可以的。先把原先的数组置为 -abs(x, x, x),省掉mark数组
while (need_todo == true) //上一次有消消乐，这次可能还需要消消乐
{
need_todo = false; //标记，看这轮需不需要消消乐
////先搞定行
for (int r = 0; r < Row; r ++)
{
for (int c = 0; c < Col - 2; c ++)
{
if ( board[r][c]!=0 && abs(board[r][c]) == abs(board[r][c+1]) && abs(board[r][c+1]) == abs(board[r][c+2]) )
{
need_todo = true;
int tmp = - abs(board[r][c]);
board[r][c] = tmp;
board[r][c+1] = tmp;
board[r][c+2] = tmp;
}
}
}
//// 再搞定列
for (int c = 0; c < Col; c ++)
{
for (int r = 0; r < Row - 2; r ++)
{
if ( board[r][c] != 0 && abs(board[r][c]) == abs(board[r+1][c]) && abs(board[r+1][c]) == abs(board[r+2][c]) )
{
need_todo = true;
int tmp = -abs(board[r][c]);
board[r][c] = tmp;
board[r+1][c] = tmp;
board[r+2][c] = tmp;
}
}
}
if (need_todo == true) //如果需要消消乐
{
//// 因为是从上往下掉落，需要一列一列的搞定。
for (int c = 0; c < Col; c ++)
{
int rr = Row - 1;
for (int r = Row - 1; r > -1; r --)
{//从下往上收
if (board[r][c] > 0)
{
board[rr][c] = board[r][c];
rr --;
}
}
while (rr > -1) //上面有空缺的，补0
{
board[rr][c] = 0;
rr --;
}
}
}
}
return board;
}
};

lc253

大于小根堆顶 可继承其会议室

class Solution {
public:
int minMeetingRooms(vector<vector<int>>& v) {
sort(v.begin(), v.end(), [](auto& a, auto& b) { return a[0] < b[0]; });
priority_queue<int, vector<int>, greater<int>> q;
for (auto& m : v) {
if (!q.empty() && m[0] >= q.top()) q.pop();
q.push(m[1]);
}
return q.size();
}
};