时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 32M,其他语言64M
思路: 嵌套循环遍历数组
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) {
for(int i = 0; i <= length - 1; i++) {
for(int j = i + 1; j <= length - 1; j++) {
if(numbers[i] == numbers[j]) {
duplication[0] = numbers[i];
return true;
}
}
}
return false;
}
}