lc3593
自底向上dfs
max dfs
cnt not need change子树
class Solution {
public:
int minIncrease(int n, vector<vector<int>>& edges, vector<int>& cost)
{
vector<vector<int>> g(n);
for (auto& e : edges) {
int x = e0, y = e1;
gx.push_back(y);
gy.push_back(x);
}
g0.push_back(-1);
int ans = 0;
auto dfs = \&(this auto&& dfs, int x, int fa) -> long long {
long long max_s = 0;
int cnt = 0;
for (int y : gx) {
if (y == fa)
continue;
long long mx = dfs(y, x);
if (mx > max_s) {
max_s = mx;
cnt = 1;
} ++else if (mx == max_s)++
++cnt++++ ;
}
++ans += gx.size() - 1 - cnt;//子树-fa-not need change++
return max_s + costx;
};
dfs(0, -1);
return ans;
}
};
lc1519
vector<int> dfs
class Solution {
public:
vector<int> countSubTrees(int n, vector<vector<int>>& edges, string labels) {
vector<int> res(n, 0);
vector<vector<int>> g(n);
for(auto& e : edges) {
ge\[0].push_back(e1);
ge\[1].push_back(e0);
}
auto dfs = \&(this auto&& dfs,int u, int p) ++-> vector<int>++
{
vector<int> cnt(26, 0);//cur
cntlabels\[u - 'a']++;
for(int v : gu) {
if(v == p) continue;
++vector<int> c = dfs(v, u);++
for(int i = 0; i < 26; i++)
cnti += ci;
//拿到的每个v 都加给cur
}
++resu = cntlabels\[u - 'a'];
//统计完v后 存入ret++
return cnt;
};
dfs(0, -1);// cur,fa
return res;
}
};
lc3254
统计连续上升的元素长度，用++一次遍历判断每个长度为k的子数组是否连续上升 借助cnt变量++
满足则记录末尾元素为能量值，否则保持-1
class Solution {
public:
vector<int> resultsArray(vector<int>& nums, int k) {
vector<int> ans(nums.size() - k + 1, -1);
int cnt = 0;
for (int i = 0; i < nums.size(); i++) {
++cnt = i == 0 || numsi == numsi - 1 + 1 ? cnt + 1 : 1;++
if (cnt >= k)
ansi - k + 1 = numsi;
}
return ans;
}
};
lc957
模拟+周期+hash
模拟每天牢房状态变化+++检测循环周期++
高效计算出 n 天后的牢房状态，避免了大数 n 的重复模拟
class Solution {
public:
vector<int> prisonAfterNDays(vector<int>& cells, int n) {
unordered_map<int, vector<int>> map; //key：第i天 value：cells状态
map.insert({0, cells});
for (int i = 1; i <= n; i++) {
if (i == 1) {
cells0 = 0;
cells7 = 0;
}
for (int j = 1; j < 7; j++) {
if (mapi - 1j - 1 == 0 && mapi - 1j + 1 == 0) cellsj = 1;
else if (mapi - 1j - 1 == 1 && mapi - 1j + 1 == 1) cellsj = 1;
else cellsj = 0;
}
map.insert({i, cells});
bool flag = true;// 判循环
for (int j = 0; j < 8; j++)
if (cellsj != map1j) flag = false;
if (i > 1 && flag) {
if (n % (i - 1) == 0) return mapi - 1; //能整除则返回上一天的状态（余数为0）
else return mapn % (i - 1);
//不能整除时返回第 n % (i - 1) 的状态
}
}
return cells;
//没出现循环直接返回模拟的结果
}
};