Leetcode 220 周赛 题解
模拟
注意一些细节,最后位置是否取值。
class Solution {
public:
string reformatNumber(string number) {
string s;
for (auto c: number)
if (c != ' ' && c != '-')
s += c;
string res;
for (int i = 0; i < s.size();) {
if ((int)s.size() - i > 4) res = res + s[i] + s[i + 1] + s[i + 2] + '-', i += 3;
else {
int x = s.size() - i;
if (x == 2) res = res + s[i] + s[i + 1];
else if (x == 3) res = res + s[i] + s[i + 1] + s[i + 2];
else res = res + s[i] + s[i + 1] + '-' + s[i + 2] + s[i + 3];
res += '-';
i += 4;
}
}
res.pop_back();
return res;
}
};
优先队列
class Solution {
public:
deque<int> q;
map<int,int> mp;
int maximumUniqueSubarray(vector<int>& nums) {
int num = 0,mn = 0;
for(int i = 0; i < nums.size(); i++){
if(!mp[nums[i]]) mp[nums[i]] = 1;
else {
while(q.size() && nums[q.front()] != nums[i]) mp[nums[q.front()]] = 0,num -= nums[q.front()],q.pop_front();
if(q.size() && nums[q.front()] == nums[i]) num -= nums[q.front()],q.pop_front();
}
q.push_back(i);
num += nums[i];
mn = max(num,mn);
}
return mn;
}
};
优先队列优化dp
\[\begin{align*}
& dp[i] = max(dp[i-1]~dp[(i-k)] + nums[i]);\\
\end{align*}
\]
& dp[i] = max(dp[i-1]~dp[(i-k)] + nums[i]);\\
\end{align*}
\]
class Solution {
public:
//dp[i] = max(dp[i-1]~dp[(i-k)] + nums[i]);
int dp[100000 + 10];
deque<int> q;
int maxResult(vector<int>& nums, int k) {
memset(dp,128,sizeof dp);
int n = nums.size();
dp[0] = nums[0];
q.push_back(0);
for(int i = 1; i < n; i ++){
while(i - q.front() > k) q.pop_front();
dp[i] = dp[q.front()] + nums[i];
while(q.size() && dp[q.back()] < dp[i]) q.pop_back();
}
return dp[n-1];
}
};
也可以用multiset优化
class Solution {
public:
int maxResult(vector<int>& nums, int k) {
int n = nums.size();
vector<int> f(n);
multiset<int> S;
f[0] = nums[0];
S.insert(f[0]);
for (int i = 1; i < n; i ++ ) {
if (i - k - 1 >= 0)
S.erase(S.find(f[i - k - 1]));
f[i] = nums[i] + *S.rbegin();
S.insert(f[i]);
}
return f[n - 1];
}
};
5632. 检查边长度限制的路径是否存在
排序 + 并查集
考虑边,对判断和原始边都进行从小到大排序,对满足限制边的原始边进行建图连边,判断是否存在/满足条件的结果。
排序后可以保证小的边限制能满足的情况下,大边同样能满足。
用并查集判断是否存在相应的边即可。
const int MAXN = 1e5 + 50;
int m, Q;
struct Node{ int u, v, w, i; } edge[MAXN], query[MAXN];
bool ans[MAXN];
bool cmp(const Node &a, const Node &b){ return a.w < b.w; }
int father[MAXN];
int getFather(int x){ return father[x] = (father[x] == x ? x: getFather(father[x])); }
void mergeFather(int x, int y){
int fx = getFather(x), fy = getFather(y);
if (fx == fy) return;
if (fx > fy) swap(fx, fy);
father[fx] = fy;
}
class Solution {
public:
vector<bool> distanceLimitedPathsExist(int n, vector<vector<int>>& edgeList, vector<vector<int>>& queries) {
m = edgeList.size(); Q = queries.size();
for (int i = 0; i < m; i++){
edge[i].u = edgeList[i][0];
edge[i].v = edgeList[i][1];
edge[i].w = edgeList[i][2];
}
for (int i = 0; i < Q; i++){
query[i].u = queries[i][0];
query[i].v = queries[i][1];
query[i].w = queries[i][2];
query[i].i = i;
}
sort(edge, edge + m, cmp); sort(query, query + Q, cmp);
for (int i = 0; i <= n; i++) father[i] = i;
for (int i = 0, k = 0; i < Q; i++){
while(k < m && edge[k].w < query[i].w) {
mergeFather(edge[k].u, edge[k].v);
k++;
}
ans[query[i].i] = (getFather(query[i].u) == getFather(query[i].v));
}
vector<bool> ret;
for (int i = 0; i < Q; i++) ret.push_back(ans[i]);
return ret;
}
};