219. Contains Duplicate II

寻找相近重复元素

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原文链接：http://www.cnblogs.com/ruruozhenhao/p/9960888.html

文章标签：

#java #python #runtime

本文探讨了如何在整数数组中查找是否存在两个不同的索引i和j，使得nums[i]=nums[j]且i与j之间的绝对差小于等于k。提供了三种不同编程语言（C++, Java, Python）的解决方案，包括详细的代码实现。

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

Approach #1: c++.

class Solution {
public:
    bool containsNearbyDuplicate(vector<int>& nums, int k) {
        unordered_map<string, int> mp;
        for (int i = 1; i <= nums.size(); ++i) {
            if ((mp[to_string(nums[i-1])] > 0) && i - mp[to_string(nums[i-1])] <= k) return true;
            else mp[to_string(nums[i-1])] = i;
        }
        return false;
    }
};

Approach #2: Java.

class Solution {
    public boolean containsNearbyDuplicate(int[] nums, int k) {
        Set<Integer> set = new HashSet<Integer>();
        for (int i = 0; i < nums.length; ++i) {
            if (i > k) set.remove(nums[i-k-1]);
            if (!set.add(nums[i])) return true;
        }
        return false;
    }
}

Approach #3: Python.

class Solution(object):
    def containsNearbyDuplicate(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: bool
        """
        dic = {}
        for i, v in enumerate(nums):
            if v in dic and i - dic[v] <= k:
                return True
            dic[v] = i
        return False