[LeetCode] 219. Contains Duplicate II 包含重复元素 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

217. Contains Duplicate 的拓展， 那道题只需判断数组中是否有重复值，而这题限制了数组中只许有一组重复的数字，而且坐标差最多k。

解法： 哈希表Hash table

Java:

class Solution {
    public boolean containsNearbyDuplicate(int[] nums, int k) {
        HashMap<Integer, Integer> index = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            if (i - index.getOrDefault(nums[i], -k - 1) <= k) return true;
            index.put(nums[i], i);
        }
        return false;
    }
}　

Python:

class Solution:
    # @param {integer[]} nums
    # @param {integer} k
    # @return {boolean}
    def containsNearbyDuplicate(self, nums, k):
        lookup = {}
        for i, num in enumerate(nums):
            if num not in lookup:
                lookup[num] = i
            else:
                # It the value occurs before, check the difference.
                if i - lookup[num] <= k:
                    return True
                # Update the index of the value.
                lookup[num] = i
        return False　　

C++:

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

　　

 

类似题目：

[LeetCode] 217. Contains Duplicate 包含重复元素

[LeetCode] 220. Contains Duplicate III 包含重复元素 III

　　　

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