求小米笔试题编程第二个

电话号码吗 #include <iostream> #include <vector> #include <string> using namespace std; int main()  { int T; cin>>T; for(int t = 0; t < T; ++t) { string str; cin>>str; int cnt[128] = {0}; for(int i = 0; i < str.length(); ++i) ++cnt[str[i]]; int num[10] = {0}; num[0] = cnt['Z']; num[2] = cnt['W']; num[4] = cnt['U']; num[6] = cnt['X']; num[8] = cnt['G']; num[1] = cnt['O'] - num[0] - num[2] - num[4]; num[3] = cnt['H'] - num[8]; num[5] = cnt['F'] - num[4]; num[7] = cnt['V'] - num[5]; num[9] = cnt['I'] - num[5] - num[6] - num[8]; for(int i = 0, j = 8; i < 10; ++i, j = (j + 1) % 10) for(int k = 0; k < num[j]; ++k) cout<<(j + 2) % 10; cout<<endl; }     return 0; }

第一组数据不一定是根节点。

//树的深度 import java.util.*; public class Main{ public static void main(String[] args) { // TODO Auto-generated method stub Scanner cin=new Scanner(System.in); int[] nn= new int[1001]; int i=0; int n=0; int count = cin.nextInt(); nn[i++] = count; while((count--)!=0) { n=cin.nextInt(); nn[i++]=n; } getlength(nn,i); } public static void getlength(int[] nn,int i){ int number = i-1; if(number==0){System.out.print(0);return;} else if(number==1){System.out.print(1);return;} else if(number>1&&number<5){System.out.print(2);return;} int start = 1; int censhu = 0; censhu = number/2; int hehe[][] = new int[censhu][2]; for(int i1=0;i1<censhu;i1++){ for(int j1=0;j1<2;j1++){ hehe[i1][j1] = nn[start]; start++; } } int maxlength1 = 2; for(int i2=0;i2<censhu-1;i2++){ for(int j2=i2+1;j2<censhu;j2++){ if(hehe[i2][1]==hehe[j2][0]){ maxlength1++; i2 = j2; } } } int maxlength2 = 2; for(int i3=1;i3<censhu-1;i3++){ for(int j3=i3+1;j3<censhu;j3++){ if(hehe[i3][1]==hehe[j3][0]){ maxlength2++; i3 = j3; } } } int maxlength = 0; if(maxlength1>maxlength2){ maxlength = maxlength1; }else{ maxlength = maxlength2; } System.out.print(maxlength); } }

亲，你用一个数组记录每个节点的父节点，然后求       每个节点到根节点的距离。

80%.....不知道错在哪里

同最多百分之20

第二题感觉我策略没错 打死只过20%

树的高度？是不是一直都只能A 10%？

不同岗位笔试题不同吧0_0

都是什么题目啊

第一个能发下么？