LeetCode 题解(196) : Contains Duplicate II

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本文介绍了一个算法，用于在给定数组中找到两个具有相同元素且索引差值不超过k的元素。提供了C++、Java和Python三种语言的实现方式。

题目：

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.

题解：

C++版：

class Solution {
public:
    bool containsNearbyDuplicate(vector<int>& nums, int k) {
        unordered_map<int, int> m;
        int index = 0;
        for(auto i : nums) {
            if(m.find(i) != m.end()) {
                if(index - m[i] <= k)
                    return true;
                else
                    m[i] = index;
            }
            m.insert(pair<int, int>(i, index));
            index++;
        }
        return false;
    }
};

Java版：

import java.util.Hashtable; 

public class Solution {
    public boolean containsNearbyDuplicate(int[] nums, int k) {
        Hashtable<Integer, Integer> m = new Hashtable<>();
        for(int i = 0; i < nums.length; i++) {
            if(m.containsKey(nums[i])) {
                if(i - m.get(nums[i]) <= k)
                    return true;
                m.put(nums[i], i);
            }
            m.put(nums[i], i);
        }
        return false;
    }
}

Python版：

class Solution(object):
    def containsNearbyDuplicate(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: bool
        """
        dict = {}
        index = 0
        for i in nums:
            if i in dict:
                if index - dict[i] <= k:
                    return True
                else:
                    dict[i] = index
            dict[i] = index
            index += 1
        return False