【leetcode】【easy】219. Contains Duplicate II

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本文详细解析了LeetCode第219题“Contains Duplicate II”的解决方案，采用滑动窗口和查找表的方法，实现高效判断数组中是否存在两个不同下标的相同元素，且下标之差不超过给定值k。代码示例清晰，讲解了两种优雅的处理方式。

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

题目链接：https://leetcode.com/problems/contains-duplicate-ii/

解题思路：滑动窗口+查找表

维护一个大小为k的查找表。

时间复杂度：O(n)

空间复杂度：O(k)

（这边对于k>nums.size()的处理不太优雅）

class Solution {
public:
    bool containsNearbyDuplicate(vector<int>& nums, int k) {
        if(k<=0) return false;
        unordered_set<int> record;
        int nlen=nums.size();
        int i=0, len=min(k,nlen);
        for(;i<len;++i){
            if(record.find(nums[i])!=record.end()) return true;
            record.insert(nums[i]);
        }
        if (len<k) return false;
        for(;i<nums.size();++i){
            if(record.find(nums[i])!=record.end()) return true;
            record.erase(nums[i-k]);
            record.insert(nums[i]);
        }
        return false;
    }
};

优雅：k>nums.size()的处理

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