﻿ 剑指offer(二十六)之数组中重复的数字 - 鸿网互联

剑指offer(二十六)之数组中重复的数字

题目描述

```<span style="font-family:SimSun;font-size:18px;">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) {
if(length<=1){
return false;
}
boolean tag=false;
for(int i=0;i<length-1;i++){
for(int j=i+1;j<length;j++){
if(numbers[i]==numbers[j]){
duplication[0]=numbers[i];
tag=true;
}
}
}
return tag;
}
}</span>```

```<span style="font-size:18px;">public class Solution {
public boolean duplicate(int numbers[],int length,int [] duplication) {

boolean tag=false;
int []temp=new int[length];
if(length<=1||numbers==null){
return tag;
}
for(int i=0;i<length;i++){
temp[numbers[i]]++;
}
for(int j=0;j<length;j++){
if(temp[j]>1){
duplication[0]=j;
tag=true;
}
}
return tag;
}
}</span>```

<