超时版：
/*Contains Duplicate II Given an array of integers and an integer k, find out whether there there are two distinct indices i and j in the array such that nums[i] = nums[j] and the difference between i and j is at most k. */ public class Leetcode2 { public static void main(String[] args) { int arr[]= {1,2,34,43,1}; System.out.println(containsNearbyDuplicate(arr,1)); // TODO Auto-generated method stub } public static boolean containsNearbyDuplicate(int[] nums, int k) { for (int i = 0; i < nums.length - 1; i++) { int temp = k + i; int y = ((temp > nums.length - 1) ? nums.length : temp+1); for (int j = 1; j < y; j++) { int x = ((i+j )> nums.length - 1) ? nums.length-1 : (i+j); if((nums[x] - nums[i])==0) return true; } } return false; } /* public static boolean containsNearbyDuplicate(int[] nums, int k) { int x; for (int i = 0; i < nums.length - 1; i++) { int temp = k + i; while(temp<=nums.length-1) { int temp2=temp; for(int j=i+1;j<temp2;j++) { x=nums[j]-nums[i]; if(x==0) return true; } } } return false; } */ }
AC版：注意集合框架的使用！！
public class Solution {
public boolean containsNearbyDuplicate(int[] nums, int k) {
Map<Integer,Integer> tm=new TreeMap<Integer,Integer>();
{
for (int i = 0; i < nums.length ; i++)
{
if(tm.containsKey(nums[i]))
{
int j=tm.get(nums[i]);
if((i-j)<=k)
return true;
}
tm.put(nums[i],i);
}
return false;
}
}
}
