找出数组中重复的数字
数组中重复的数字
http://www.nowcoder.com/questionTerminal/623a5ac0ea5b4e5f95552655361ae0a8
思路分析:
如果要求找到的是第一个重复的数字 在下面代码里for循环中要用if不能用while
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++){
if(i != numbers[i]){
if(numbers[i] == numbers[numbers[i]]){
duplication[0] = numbers[i];
return true;
}
//开始进行交换操作
int tmp = numbers[i];
numbers[i] = numbers[tmp];
numbers[tmp] = tmp;
}
}
return false;
}
} 通过hashMap表来实现
import java.util.*;
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;
}
}
Map hashMap = new HashMap(length);
for(int i = 0; i < length; i++){
if(hashMap.containsKey(numbers[i])){
duplication[0] = numbers[i];
return true;
}else{
hashMap.put(numbers[i], i);
}
}
return false;
}
}