class Solution {
public:
vector<vector<int>> findWinners(vector<vector<int>>& matches) {
}
};
2225. 找出输掉零场或一场比赛的玩家
给你一个整数数组 matches 其中 matches[i] = [winneri, loseri] 表示在一场比赛中 winneri 击败了 loseri 。
返回一个长度为 2 的列表 answer ：
answer[0] 是所有 没有 输掉任何比赛的玩家列表。answer[1] 是所有恰好输掉 一场 比赛的玩家列表。两个列表中的值都应该按 递增 顺序返回。
注意：
示例 1：
输入：matches = [[1,3],[2,3],[3,6],[5,6],[5,7],[4,5],[4,8],[4,9],[10,4],[10,9]] 输出：[[1,2,10],[4,5,7,8]] 解释： 玩家 1、2 和 10 都没有输掉任何比赛。 玩家 4、5、7 和 8 每个都输掉一场比赛。 玩家 3、6 和 9 每个都输掉两场比赛。 因此，answer[0] = [1,2,10] 和 answer[1] = [4,5,7,8] 。
示例 2：
输入：matches = [[2,3],[1,3],[5,4],[6,4]] 输出：[[1,2,5,6],[]] 解释： 玩家 1、2、5 和 6 都没有输掉任何比赛。 玩家 3 和 4 每个都输掉两场比赛。 因此，answer[0] = [1,2,5,6] 和 answer[1] = [] 。
提示：
1 <= matches.length <= 105matches[i].length == 21 <= winneri, loseri <= 105winneri != loserimatches[i] 互不相同原站题解
php 解法, 执行用时: 652 ms, 内存消耗: 98.4 MB, 提交时间: 2024-05-22 09:32:35
class Solution {
/**
* @param Integer[][] $matches
* @return Integer[][]
*/
function findWinners($matches) {
$hash = [];
foreach ($matches as $mat) {
if (!isset($hash[$mat[0]])) $hash[$mat[0]] = 0;
$hash[$mat[1]]++;
}
ksort($hash);
$res = [[],[]];
foreach ($hash as $k => $v) {
if ($v == 0) {
$res[0][] = $k;
} else if ($v == 1) {
$res[1][] = $k;
}
}
return $res;
}
}
rust 解法, 执行用时: 65 ms, 内存消耗: 9.7 MB, 提交时间: 2024-05-22 09:29:49
use std::collections::HashMap;
impl Solution {
pub fn find_winners(matches: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
let mut loss_count = HashMap::new();
for m in matches {
loss_count.entry(m[0]).or_insert(0);
*loss_count.entry(m[1]).or_insert(0) += 1;
}
let mut ans = vec![vec![], vec![]];
for (player, cnt) in loss_count {
if cnt < 2 {
ans[cnt as usize].push(player);
}
}
ans[0].sort_unstable();
ans[1].sort_unstable();
ans
}
}
javascript 解法, 执行用时: 348 ms, 内存消耗: 90.6 MB, 提交时间: 2024-05-22 09:29:20
/**
* @param {number[][]} matches
* @return {number[][]}
*/
var findWinners = function(matches) {
const lossCount = new Map();
for (const [winner, loser] of matches) {
if (!lossCount.has(winner)) {
lossCount.set(winner, 0);
}
lossCount.set(loser, (lossCount.get(loser) ?? 0) + 1);
}
const ans = [[], []];
for (const [player, cnt] of lossCount) {
if (cnt < 2) {
ans[cnt].push(player);
}
}
ans[0].sort((a, b) => a - b);
ans[1].sort((a, b) => a - b);
return ans;
};
cpp 解法, 执行用时: 403 ms, 内存消耗: 160.8 MB, 提交时间: 2024-05-22 09:28:59
class Solution {
public:
vector<vector<int>> findWinners(vector<vector<int>>& matches) {
unordered_map<int, int> loss_count;
for (auto& m : matches) {
if (!loss_count.contains(m[0])) {
loss_count[m[0]] = 0;
}
loss_count[m[1]]++;
}
vector<vector<int>> ans(2);
for (auto& [player, cnt] : loss_count) {
if (cnt < 2) {
ans[cnt].push_back(player);
}
}
ranges::sort(ans[0]);
ranges::sort(ans[1]);
return ans;
}
};
java 解法, 执行用时: 67 ms, 内存消耗: 90.6 MB, 提交时间: 2024-05-22 09:28:31
public class Solution {
public List<List<Integer>> findWinners(int[][] matches) {
Map<Integer, Integer> lossCount = new HashMap<>();
for (int[] m : matches) {
if (!lossCount.containsKey(m[0])) {
lossCount.put(m[0], 0);
}
lossCount.merge(m[1], 1, Integer::sum);
}
List<List<Integer>> ans = List.of(new ArrayList<>(), new ArrayList<>());
for (Map.Entry<Integer, Integer> e : lossCount.entrySet()) {
int cnt = e.getValue();
if (cnt < 2) {
ans.get(cnt).add(e.getKey());
}
}
Collections.sort(ans.get(0));
Collections.sort(ans.get(1));
return ans;
}
}
python3 解法, 执行用时: 250 ms, 内存消耗: 60.2 MB, 提交时间: 2024-05-22 09:28:05
class Solution:
def findWinners(self, matches: List[List[int]]) -> List[List[int]]:
players = set(x for m in matches for x in m)
loss_count = Counter(loser for _, loser in matches)
return [sorted(x for x in players if x not in loss_count),
sorted(x for x, c in loss_count.items() if c == 1)]
javascript 解法, 执行用时: 384 ms, 内存消耗: 83.1 MB, 提交时间: 2022-11-29 15:58:51
/**
* @param {number[][]} matches
* @return {number[][]}
*/
var findWinners = function(matches) {
let wins = new Map();
let loses = new Map();
let ansWin = [];
let ansLose = [];
for(let [win, lose] of matches) {
wins.set(win, (wins.get(win) || 0) + 1)
loses.set(lose, (loses.get(lose) || 0) + 1)
}
for(let [key, val] of loses.entries()) {
if(val === 1) {
ansLose.push(key)
}
}
for(let [key, val] of wins.entries()) {
if(!loses.has(key)) {
ansWin.push(key)
}
}
ansWin.sort((a, b) => a - b);
ansLose.sort((a, b) => a - b);
return [ansWin, ansLose]
};
golang 解法, 执行用时: 376 ms, 内存消耗: 25.8 MB, 提交时间: 2022-11-29 15:57:51
func findWinners(matches [][]int) [][]int {
man := map[int]struct{}{}
cnt := map[int]int{}
for _, p := range matches {
man[p[0]] = struct{}{}
man[p[1]] = struct{}{} // 记录所有出现过的人
cnt[p[1]]++ // 统计每个人输的次数
}
ans := [][]int{{}, {}}
for v := range man {
if c := cnt[v]; c < 2 {
ans[c] = append(ans[c], v)
}
}
sort.Ints(ans[0])
sort.Ints(ans[1])
return ans
}
python3 解法, 执行用时: 316 ms, 内存消耗: 54.2 MB, 提交时间: 2022-11-29 15:57:27
class Solution:
def findWinners(self, matches: List[List[int]]) -> List[List[int]]:
freq = Counter()
for winner, loser in matches:
if winner not in freq:
freq[winner] = 0
freq[loser] += 1
ans = [[], []]
for key, value in freq.items():
if value < 2:
ans[value].append(key)
ans[0].sort()
ans[1].sort()
return ans