Given typedText containing uppercase and lowercase English letters, return upper_count - lower_count.
Example: typedText = "AbCd" → upper = 2, lower = 2, answer = 0.
解法
单次遍历，分别统计大写和小写字母数后相减。复杂度 O(n)。
def letter_case_diff(s: str) -> int:
return sum(1 for c in s if c.isupper()) - sum(1 for c in s if c.islower())class Solution {
static int letterCaseDiff(String s) {
int u = 0, l = 0;
for (char c : s.toCharArray()) {
if (Character.isUpperCase(c)) u++;
else if (Character.isLowerCase(c)) l++;
}
return u - l;
}
}#include <string>
#include <cctype>
using namespace std;
int letterCaseDiff(string s) {
int u = 0, l = 0;
for (char c : s) {
if (isupper((unsigned char)c)) u++;
else if (islower((unsigned char)c)) l++;
}
return u - l;
}A bird builds a nest by collecting sticks in a forest. forest[] non-negative integers; non-zero = stick of that length, zero = empty cell. Start at index bird, fly right searching the next non-zero cell, take it (mark zero), append the stick to the nest. Then flip direction (right → left → right ...). Continue until total nest length ≥ 100. Return 0-based indices of sticks collected, in order.
Example: forest = [3, 0, 50, 0, 80, 0, 25], bird = 0 → fly right, pick 3 at index 0, total 3; flip left → no stick; flip right → pick 50 at index 2, total 53; flip left → no stick; flip right → pick 80, total 133 ≥ 100, stop.
解法
模拟：维护 pos 和 direction；前进直到非零格收走，翻转方向，循环直到总和 ≥ 100 或越界。复杂度 O(n)。
def bird_sticks(forest: list[int], bird: int) -> list[int]:
f = list(forest)
n = len(f)
out, total = [], 0
pos, d = bird, 1
while total < 100:
i = pos
while 0 <= i < n and f[i] == 0:
i += d
if i < 0 or i >= n:
break
out.append(i)
total += f[i]
f[i] = 0
pos, d = i + d, -d
return outimport java.util.*;
class Solution {
static int[] birdSticks(int[] forest, int bird) {
int n = forest.length, total = 0, pos = bird, dir = 1;
List<Integer> out = new ArrayList<>();
while (total < 100) {
int i = pos;
while (i >= 0 && i < n && forest[i] == 0) i += dir;
if (i < 0 || i >= n) break;
out.add(i);
total += forest[i];
forest[i] = 0;
pos = i + dir;
dir = -dir;
}
int[] arr = new int[out.size()];
for (int k = 0; k < arr.length; k++) arr[k] = out.get(k);
return arr;
}
}#include <vector>
using namespace std;
vector<int> birdSticks(vector<int> forest, int bird) {
int n = forest.size(), total = 0, pos = bird, dir = 1;
vector<int> out;
while (total < 100) {
int i = pos;
while (i >= 0 && i < n && forest[i] == 0) i += dir;
if (i < 0 || i >= n) break;
out.push_back(i);
total += forest[i];
forest[i] = 0;
pos = i + dir;
dir = -dir;
}
return out;
}Events of two types:
MESSAGE id<num> <timestamp> <mention_text>—mention_textis space-separated tokens ofid<num>,ALL, orHERE.OFFLINE id<num>— that user becomes inactive (still exists, butHEREwon't count them).
ALL mentions every existing user; HERE mentions every currently-active user. Each mention token counted once per message regardless of repeats; self-mentions count. Return strings "id<num>=<count>" sorted lexicographically.
解法
维护两个集合 all_users、active_users 和计数表。每条 MESSAGE：把发送者放入两个集合，构造去重 mention 集合，每个成员计数加 1。OFFLINE：仅从 active_users 移除。复杂度 O(total events + total mentions)。
def chat_mentions(events: list[str]) -> list[str]:
all_, active = set(), set()
cnt = {}
for ev in events:
parts = ev.split()
if parts[0] == "MESSAGE":
sender = parts[1]
all_.add(sender); active.add(sender)
mentioned = set()
for tok in parts[3:]:
if tok == "ALL": mentioned |= all_
elif tok == "HERE": mentioned |= active
else: mentioned.add(tok)
for m in mentioned:
cnt[m] = cnt.get(m, 0) + 1
else:
active.discard(parts[1])
return [f"{k}={v}" for k, v in sorted(cnt.items())]import java.util.*;
class Solution {
static List<String> chatMentions(List<String> events) {
Set<String> all = new HashSet<>(), active = new HashSet<>();
Map<String, Integer> cnt = new TreeMap<>();
for (String ev : events) {
String[] parts = ev.split("\\s+");
if (parts[0].equals("MESSAGE")) {
String sender = parts[1];
all.add(sender); active.add(sender);
Set<String> mentioned = new HashSet<>();
for (int i = 3; i < parts.length; i++) {
if (parts[i].equals("ALL")) mentioned.addAll(all);
else if (parts[i].equals("HERE")) mentioned.addAll(active);
else mentioned.add(parts[i]);
}
for (String m : mentioned) cnt.merge(m, 1, Integer::sum);
} else {
active.remove(parts[1]);
}
}
List<String> out = new ArrayList<>();
for (var e : cnt.entrySet()) out.add(e.getKey() + "=" + e.getValue());
return out;
}
}#include <vector>
#include <string>
#include <sstream>
#include <unordered_set>
#include <map>
using namespace std;
vector<string> chatMentions(vector<string>& events) {
unordered_set<string> all_, active;
map<string, int> cnt;
for (auto& ev : events) {
stringstream ss(ev);
vector<string> parts;
string tok;
while (ss >> tok) parts.push_back(tok);
if (parts[0] == "MESSAGE") {
const string& sender = parts[1];
all_.insert(sender); active.insert(sender);
unordered_set<string> mentioned;
for (int i = 3; i < (int)parts.size(); i++) {
if (parts[i] == "ALL") mentioned.insert(all_.begin(), all_.end());
else if (parts[i] == "HERE") mentioned.insert(active.begin(), active.end());
else mentioned.insert(parts[i]);
}
for (auto& m : mentioned) cnt[m]++;
} else {
active.erase(parts[1]);
}
}
vector<string> out;
for (auto& [k, v] : cnt) out.push_back(k + "=" + to_string(v));
return out;
}