一 分别比较
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(numbers == null || length == 0){
return false;
}
for(int i = 0;i < length;i++){
if(numbers[i] < 0 || numbers[i] > length-1){
return false;
}
}
for(int i = 0;i < length;i++){
for(int j = i + 1;j < length;j++){
if(numbers[i] == numbers[j]){
duplication[0] = numbers[i];
return true;//返回第一个
}
}
}
return false;
}
}
二 hashSet做法
import java.util.HashSet;
import java.util.Set;
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(numbers == null || length == 0){
return false;
}
for(int i = 0;i < length;i++){
if(numbers[i] < 0 || numbers[i] > length){
return false;
}
}
Set<Integer> set = new HashSet<Integer>();
for(int i = 0;i < length;i++){
if(!set.add()){
duplication[0] = numbers[i];
return true;
}
}
return false;
}
}
三 不借助任何(用数组下标的关系来做)
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(numbers == null || length == 0){
return false;
}
for(int i = 0;i < length;i++){
if(numbers[i] < 0 || numbers[i] > length){
return false;
}
}
for(int i = 0;i < length;i++){
while(numbers[i] != i){
if(numbers[i] == numbers[numbers[i]]){
duplication[0] = numbers[i];
return true;
}
int temp = numbers[i];
numbers[i] = numbers[temp];
numbers[temp] = temp;
}
}
return false;
}
}
四 鸽巢原理(不提供代码)