第 107 场LeetCode双周赛
A 最大字符串配对数目
显然各字符串对 间匹配的先后顺序不影响最大匹配数目, 可以从后往前遍历数组, 判断前面是否有和当前末尾构成匹配的.
class Solution {
public:
int maximumNumberOfStringPairs(vector<string> &words) {
int res = 0;
while (words.size() > 1) {
auto &s = words.back();
reverse(s.begin(), s.end());
for (int i = words.size() - 2; i >= 0; i--)
if (s == words[i]) {
res++;
break;
}
words.pop_back();
}
return res;
}
};
B 构造最长的新字符串
记忆化搜索: 定义状态 p a a , b b , a b , l a s t p_{aa,bb,ab,last} paa,bb,ab,last为剩余三种字符串分别为aa、bb、ab个, 且上一位为last(1代表’A’, 2代表’B’)情况下, 后面可以生成的最长字符串的长度, 枚举可用的三种字符串即可.
class Solution {
public:
int longestString(int x, int y, int z) {
int p[x + 1][y + 1][z + 1][3];
memset(p, -1, sizeof(p));
function<int(int, int, int, int)> get = [&](int aa, int bb, int ab, int last) {//last: 1->A 2->B
if (p[aa][bb][ab][last] != -1)
return p[aa][bb][ab][last];
p[aa][bb][ab][last] = 0;
if (aa && last != 1)
p[aa][bb][ab][last] = max(p[aa][bb][ab][last], 2 + get(aa - 1, bb, ab, 1));
if (bb && last != 2)
p[aa][bb][ab][last] = max(p[aa][bb][ab][last], 2 + get(aa, bb - 1, ab, 2));
if (ab && last != 1)
p[aa][bb][ab][last] = max(p[aa][bb][ab][last], 2 + get(aa, bb, ab - 1, 2));
return p[aa][bb][ab][last];
};
return get(x, y, z, 0);// last!=1,2 (初始情况没有上一位)
}
};
C 字符串连接删减字母
动态规划: 定义 p i , f r o n t , b a c k p_{i,front,back} pi,front,back为生成的首位为 f r o n t front front末位为 b a c k back back的 s t r i str_i stri的最长长度, 若 p i , f r o n t , b a c k p_{i,front,back} pi,front,back状态是可达的, 有两种方式对 p i + 1 , f o n t ′ , b a c k ′ p_{i+1,font',back'} pi+1,font′,back′进行更新(对应题中两种操作).
class Solution {
public:
int minimizeConcatenatedLength(vector<string> &li) {
int n = li.size();
int p[n][26][26];//i,front,back
for (int i = 0; i < n; i++)
for (int j = 0; j < 26; j++)
for (int k = 0; k < 26; k++)
p[i][j][k] = INT32_MAX;// 标志状态不可达
p[0][li[0][0] - 'a'][li[0].back() - 'a'] = li[0].size();
for (int i = 1; i <= n - 1; i++) {
for (int front = 0; front < 26; front++) {
for (int back = 0; back < 26; back++)
if (p[i - 1][front][back] != INT32_MAX) {
p[i][front][li[i].back() - 'a'] = min(p[i][front][li[i].back() - 'a'], p[i - 1][front][back] + (int) (li[i][0] - 'a' == back ? li[i].size() - 1 : li[i].size()));// join(str_(i-1) , li[i])
p[i][li[i][0] - 'a'][back] = min(p[i][li[i][0] - 'a'][back], p[i - 1][front][back] + (int) (li[i].back() - 'a' == front ? li[i].size() - 1 : li[i].size())); //join(li[i], str_(i-1))
}
}
}
int res = INT32_MAX;
for (int front = 0; front < 26; front++)
for (int back = 0; back < 26; back++)
res = min(res, p[n - 1][front][back]);
return res;
}
};
D 统计没有收到请求的服务器数目
离线查询+双指针: 把日志和查询放在一个有序表 l i li li中, 并按时间非降序排序, 同时保证相同时间的日志在查询之前. 之后用双指针的方法维护 l i li li中一段时间差最大为x的区间, 及该区间上不同的服务器个数为 c n t cnt cnt, 每次移动右指针遇到是查询时, 更新左指针, 对应查询的答案即为 n − c n t n-cnt n−cnt.
class Solution {
public:
vector<int> countServers(int n, vector<vector<int>> &logs, int x, vector<int> &queries) {
vector<tuple<int, int, int>> li;// time, type(0:log ,1: query), server_id/query_index
for (auto &i: logs)
li.emplace_back(i[1], 0, i[0]);// time, type, server_id
int m = queries.size();
for (int i = 0; i < m; i++)
li.emplace_back(queries[i], 1, i);// time, type, query_index
sort(li.begin(), li.end());
unordered_map<int, int> cnt_server;//cnt_server[i] 服务器i当前出现次数
int cnt = 0;
vector<int> res(m);
for (int l = 0, r = 0; r < li.size(); r++) {//遍历li[r]
if (get<1>(li[r]) == 0) {//log
int server_id = get<2>(li[r]);
if (cnt_server[server_id]++ == 0)
cnt++;
} else {//query
int time_cur = get<0>(li[r]), query_index = get<2>(li[r]);
for (; get<0>(li[l]) < time_cur - x; l++) {//更新l
if (get<1>(li[l]) == 0) {//log
if (--cnt_server[get<2>(li[l])] == 0)// 维护cnt_server
cnt--;
}
}
res[query_index] = n - cnt;
}
}
return res;
}
};