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; } }