字典树

今日头条  异或

题目描述

给定整数m以及n各数字A1,A2,..An，将数列A中所有元素两两异或，共能得到n(n-1)/2个结果，请求出这些结果中大于m的有多少个。

输入描述:

第一行包含两个整数n,m. 

第二行给出n个整数A1，A2，...，An。

数据范围

对于30%的数据，1 <= n, m <= 1000

对于100%的数据，1 <= n, m, Ai <= 10^5

输出描述:

输出仅包括一行，即所求的答案

示例1

输入

3 10  
6 5 10

输出

2

思路：用整数的位来构造字典树。

import java.util.*;
public class Main{
    static class TireTree{
        TireTree[] next = new TireTree[2];
        int count = 1;
    }
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        sc.nextLine();
        int[] a = new int[n];
        for(int i = 0;i<n;i++){
            a[i] = sc.nextInt();
        }
        System.out.println(helper(a,m));
    }
    
    public static long helper(int[] a,int m){
        TireTree tree = buildTree(a);
        long result = 0;
        for(int i = 0;i<a.length;i++){
            result+=slove(a[i],m,tree,31);
        }
        return result/2;
    }
    
    public static long slove(int num,int m,TireTree tree,int index){
        if(tree == null) return 0;
        int aDight = (num>>index)&1;
        int mDight = (m>>index)&1;
        if(aDight == 1&&mDight == 1){
            return slove(num,m,tree.next[0],index-1);
        }else if(aDight == 0&&mDight == 1){
            return slove(num,m,tree.next[1],index-1);
        }else if(aDight == 1&&mDight == 0){
            long p = tree.next[0]==null?0:tree.next[0].count;
            long q = slove(num,m,tree.next[1],index-1);
            return p+q;
        }else{
            long p = tree.next[1] == null?0:tree.next[1].count;
            long q = slove(num,m,tree.next[0],index-1);
            return p+q;
        }
    }
    
    public static TireTree buildTree(int[] a){
        TireTree tree = new TireTree();
        for(int i = 0;i<a.length;i++){
            TireTree cur = tree;
            for(int j = 31;j>=0;j--){
                int d = (a[i]>>j)&1;
                if(cur.next[d] == null){
                    cur.next[d] = new TireTree();
                }else{
                    cur.next[d].count++;
                }
                cur = cur.next[d];
            }
        }
        return tree;
    }
}

440. K-th Smallest in Lexicographical Order

 思路：以当前数字开头的数的个数和k比较

class Solution {
    public int findKthNumber(int n, int k) {
        int cur = 1;
        k--;
        while(k>0){
            int len = helper(n,cur);
            if(len>k){
                k--;
                cur*=10;
            }else{
                cur++;
                k-=len;
            }
        }
        return cur;
    }
    
    public int helper(int n,long n1){
        int len = 0;
        long n2 = n1+1;
        while (n1 <= n) {
            len += Math.min(n + 1, n2) - n1;
            n1 *= 10;
            n2 *= 10;
        }
        return len;
    }
}

212. Word Search II

 

class Solution {
    public List<String> findWords(char[][] board, String[] words) {
        List<String> list = new ArrayList<>();
        TrieNode root = buildTree(words);
        for(int i = 0;i<board.length;i++){
            for(int j = 0;j<board[0].length;j++){
                helper(board,i,j,root,list);
            }
        }
        return list;
    }
    
    public void helper(char[][] board,int i,int j,TrieNode root,List<String> list){
        if(i<0||j<0||i==board.length||j==board[0].length) return;
        int index = board[i][j] - 'a';
        if(index>25||index<0) return;
        if(root.next[index] == null) return;
        root = root.next[index];
        if(root.str!=null){
            list.add(root.str);
            root.str = null;
        }
        board[i][j]^=256;
        helper(board,i-1,j,root,list);
        helper(board,i,j-1,root,list);
        helper(board,i+1,j,root,list);
        helper(board,i,j+1,root,list);
        board[i][j]^=256;
    }
    
    public TrieNode buildTree(String[] words){
        TrieNode root = new TrieNode();
        for(String s:words){
            TrieNode p = root;
            for(char c:s.toCharArray()){
                int index = c-'a';
                if(p.next[index] == null) p.next[index] = new TrieNode();
                p = p.next[index];
            }
            p.str = s;
        }
        return root;
    }

    
}

class TrieNode{
    TrieNode[] next;
    String str;
    public TrieNode(){
        next = new TrieNode[26];
    }
}