题目一:找出数组中重复的数字
法一:
import java.util.HashMap;
public class Solution {
// Parameters:
// numbers: an array of integers
// length: the length of array numbers
// duplication: (Output) the duplicated number in the array number,length of duplication array is 1,so using duplication[0] = ? in implementation;
// Here duplication like pointor in C/C++, duplication[0] equal *duplication in C/C++
// 这里要特别注意~返回任意重复的一个,赋值duplication[0]
// Return value: true if the input is valid, and there are some duplications in the array number
// otherwise false
public boolean duplicate(int numbers[],int length,int [] duplication) {
HashMap<Integer,Integer> map=new HashMap<Integer,Integer>();
boolean flag=false;
if(length==0||numbers==null){
duplication[0]=-1;
return flag;
}
for(Integer num: numbers){
if(map.containsKey(num)){
duplication[0]=num;
flag=true;
break;
}else{
map.put(num,0);
}
}
return flag;
}
}
法二:
import java.util.Arrays;
public class Solution {
// Parameters:
// numbers: an array of integers
// length: the length of array numbers
// duplication: (Output) the duplicated number in the array number,length of duplication array is 1,so using duplication[0] = ? in implementation;
// Here duplication like pointor in C/C++, duplication[0] equal *duplication in C/C++
// 这里要特别注意~返回任意重复的一个,赋值duplication[0]
// Return value: true if the input is valid, and there are some duplications in the array number
// otherwise false
public boolean duplicate(int numbers[],int length,int [] duplication) {
boolean flag=false;
if(numbers==null||length<0)
return flag;
Arrays.sort(numbers);
for(int i=0;i<length-1;i++){
if(numbers[i]==numbers[i+1]){
duplication[0]=numbers[i];
flag=true;
break;
}
}
return flag;
}
}