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 == 0) return false; for(int i = 0; i < length; i++) { int index = numbers[i]; if(index < 0) { index = numbers[i] + length; } if(numbers[index] < 0){ duplication[0] = index; return true; } numbers[index] -= length; } return false; } }
原文:https://www.cnblogs.com/yihangZhou/p/10503234.html