58笔试 简单回忆版 （2022/03/12 20:30结束 三道easy题，第二题我傻逼了没交上去。。。）

1.最长不重复数组

    int solve(int[] arr) {
        Queue<Integer> queue = new LinkedList<>();
        int ans = 0;
        for (int i = 0; i < arr.length; i++) {
            queue.offer(arr[i]);
            while (queue.contains(arr[i])) {
                queue.poll();
            }
            ans = Math.max(queue.size(), ans);
        }
        return ans;

    }

在一堆里找不重复且平均值相同的数组

 ArrayList<ArrayList<Integer>>  list = new ArrayList<>();
    HashSet<Integer> set = new HashSet<>();
    boolean[] st;
    public ArrayList<ArrayList<Integer>>  process (int[] allScoreArray, int examCount, int lastScore) {
        // write code here
        st = new boolean[allScoreArray.length];
        for(int i = 1; i < allScoreArray.length; i++){
            if(allScoreArray[i] == allScoreArray[i-1]){
                st[i] = true;
            }
        }
        Arrays.sort(allScoreArray);
        dfs(allScoreArray,examCount,lastScore,0);
        return list;
    }
    ArrayList<Integer> tmp = new ArrayList<>();

    void dfs(int[] allScoreArray, int examCount, int la,int n){
        if(tmp.size()!= 0 && n%tmp.size()!=0){
            return;
        }
        if(tmp.size() == examCount){
            if(n%examCount== 0 && tmp.get(examCount-1) == la ){
                ArrayList<Integer> t = new ArrayList<>(tmp);
                list.add(t);
                return;
            }
        }
        int cnt = 0;
        for(Integer i:allScoreArray){
            if(!st[cnt++] && !set.contains(i)){
                set.add(i);
                tmp.add(i);
                dfs(allScoreArray,examCount,la,n+i);
                set.remove(i);
                tmp.remove(i);
            }
        }
    }

跳箱子升级版

int solve(int k) {
        int[] dp = new int[50];
        dp[1] = 1;

        for (int i = 1; i <= 40; i++) {
            if (i % 2 == 0) {
                dp[i + 1] = dp[i] + 1;
                dp[i + 2] = dp[i] + 2;
            } else {
                dp[i + 1] = dp[i] + 1;
                dp[i + 3] = dp[i] + 1;
            }
        }
        return dp[k];
    }