[LeetCode 219] Contains Duplicate II

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本文介绍了一个算法，用于在给定数组和整数k的情况下，判断是否存在两个不同的索引，使得它们对应的元素相等且索引之差不超过k。使用哈希表实现O(n)的时间复杂度。

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 difference between i and j is at most k.

solution:

Use hashmap, key is the element value, value is the element index.

O(n)

public boolean containsNearbyDuplicate(int[] nums, int k) {
        int len = nums.length;
        if(len <= 1) return false;
        Map<Integer, Integer> count = new HashMap<>();
        for(int i = 0;i < len; i++) {
            if(count.containsKey(nums[i])) {
                if(Math.abs(count.get(nums[i]) - i) <= k) return true;
            }
            count.put(nums[i], i);
        }
        return false;
    }