219. Contains Duplicate II

Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k.

Example 1:

Input: nums = [1,2,3,1], k = 3
Output: true

Example 2:

Input: nums = [1,0,1,1], k = 1
Output: true

Example 3:

Input: nums = [1,2,3,1,2,3], k = 2
Output: false

class Solution {
    public boolean containsNearbyDuplicate(int[] nums, int k) {
        boolean result = false;
        for(int i = 0; i < nums.length - 1; i++){
            for(int j = i+1; j < nums.length; j++){
                if((nums[i]==nums[j]) && Math.abs(i - j) <= k) result = true;
            }
        }
        return result;
    }
}

1。 懒人方法，双重循环

class Solution {
    public boolean containsNearbyDuplicate(int[] nums, int k) {
        // boolean result = false;
        // for(int i = 0; i < nums.length - 1; i++){
        //     for(int j = i+1; j < nums.length; j++){
        //         if((nums[i]==nums[j]) && Math.abs(i - j) <= k) result = true;
        //     }
        // }
        // return result;

HashMap<Integer, Integer> hashMap = new HashMap<Integer, Integer>();

for (int i = 0; i < nums.length; i++) {
if (hashMap.containsKey(nums[i]) && i - hashMap.get(nums[i]) <= k) {
return true;
}
hashMap.put(nums[i], i);
}

return false;

    }
}

方法2：用hashmap来读和存数，遇到相同的计算一下距离。注意题目是只要存在有两个相同的，而且index之差小于等于k即可，一经成立就可以返回true。

https://my.oschina.net/Tsybius2014/blog/517511