Leetcode Solution

久违的空闲周末, 久违的周赛
昨天晚上把键盘ESC和Caps互相映射了一下, 导致今天打码卡手, 烦

总结

1. 用java做周赛的话, 就有点太慢了, 两部分原因: 记不住; 语法罗嗦
   以后写代码多记忆一下; 试试scala
2. PriorityQueue用法, 最好自己写个Pair:

PriorityQueue<Pair> que = new PriorityQueue<>();
que.poll();
que.add();

static class Pair {
    @Override
    public int compareTo(Pair p) {
        return Integer.compare(x, p.x);
    }
} 

1. 以后所有的排序就按stream+方法/lambda的方式吧

map.entrySet().stream()
    .sorted(this::compare)
    .map(Map.Entry::getKey)
    .collect(Collectors.toList());

1. 多用用map.computeIfAbsent, map.computeIfPresent方法

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5303. Decrypt String from Alphabet to Integer Mapping

## 5303. Decrypt String from Alphabet to Integer Mapping ### 思路 水题不写思路了 ### 要点 无 ### 代码 ```java class Solution { public String freqAlphabets(String s) { StringBuilder builder = new StringBuilder(); for (int i=0; i= s.length()) builder.append((char)(s.charAt(i)-'1'+'a')); else { if (s.charAt(i+2)=='#') { builder.append((char)((s.charAt(i)-'0')*10+s.charAt(i+1)-'1'+'a')); i += 2; } else builder.append((char)(s.charAt(i)-'1'+'a')); } } return builder.toString(); } } ``` ();>

5304. XOR Queries of a Subarray

## 5304. XOR Queries of a Subarray ### 思路 异或前缀和 ### 要点 注意处理下标越界两种方式: 条件判断, 重定义数组大小 简单思考一下就行, 拿不准就条件判断, 免得浪费时间. ### 代码 ```java class Solution { public int[] xorQueries(int[] arr, int[][] queries) { int[] sum = new int[arr.length+1]; sum[1] = arr[0]; for (int i=2; i<=arr.length; i++) { sum[i] = sum[i-1] ^ arr[i-1]; System.out.println(sum[i]); } int size = 0; int[] ans = new int[queries.length]; for (int[] q: queries) ans[size++] = sum[q[1]+1] ^ sum[q[0]]; return ans; } } ```

5305. Get Watched Videos by Your Friends

## 5305. Get Watched Videos by Your Friends ### 思路 脑子卡壳, 首先想了个错的dfs思路, 然后WA, 最后还是安心写最短路了-_- dijkstra按id求个最短路dist[], 然后Map记录, 最后排序就行. 其实还是个水题, 但是java写起来很卡手, 有些类记不起来, 得练练. ### 要点 1. PriorityQueue用法, 最好自己写个Pair: ```java PriorityQueue que = new PriorityQueue<>(); que.poll(); que.add(); static class Pair { @Override public int compareTo(Pair p) { return Integer.compare(x, p.x); } } ``` 2. 以后所有的排序就按stream+方法/lambda的方式吧 ``` map.entrySet().stream() .sorted(this::compare) .map(Map.Entry::getKey) .collect(Collectors.toList()); ``` 3. 多用用`map.computeIfAbsent`, `map.computeIfPresent`方法 ### 代码 ```java class Solution { private int[][] next; private List> elem; public List watchedVideosByFriends(List> watchedVideos, int[][] friends, int id, int level) { this.next = friends; this.elem = watchedVideos; Map map = new HashMap<>(); int[] dis = new int[next.length]; dij(id, dis); for (int i = 0; i map, int id) { for (String s: elem.get(id)) { map.computeIfAbsent(s, x -> 0); map.computeIfPresent(s, (k, v) -> v+1); } } private int compare(Map.Entry x, Map.Entry y) { Integer xv = x.getValue(), yv = y.getValue(); if (xv.equals(yv)) return x.getKey().compareTo(y.getKey()); return xv.compareTo(yv); } private void dij(int start, int[] dis) { PriorityQueue que = new PriorityQueue<>(); Arrays.fill(dis, Integer.MAX_VALUE); dis[start] = 0; que.add(new Pair(start, 0)); while (!que.isEmpty()) { Pair from = que.poll(); if (from.second != dis[from.first]) continue; // System.out.println(from.first + ", " + from.second); for (int to: next[from.first]) { if (dis[to] < dis[from.first] + 1) continue; dis[to] = dis[from.first] + 1; que.add(new Pair(to, dis[to])); } } } static class Pair implements Comparable { public int first, second; public Pair(int first, int second) { this.first = first; this.second = second; } @Override public int compareTo(Pair o) { return Integer.compare(second, o.second); } } } ``` ,>,>;>,>