leetcode 217|219|220 contains duplicate 1|2|3

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本文介绍了三种查找数组中重复元素的技术：通过排序去重、使用哈希表（map）和处理接近重复项的特殊情况。提供了详细的算法实现，并讨论了不同方法的优缺点。

217

Given an array of integers, find if the array contains any duplicates. Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct.

way-1:排序去重看长度
way-2:用map来存

class Solution {
public:
    bool containsDuplicate(vector<int>& nums) 
    {
        //way-1
        /*
        sort(nums.begin(), nums.end());
        int first = nums.size();
        auto it = unique(nums.begin(), nums.end());
        return (it-nums.begin()) != first;
        */
        
        //way-2
        map<int,int> mm;
        for (int i = 0; i < nums.size(); i++)
        {
            if (mm.find(nums[i]) == mm.end())
                mm[nums[i]]++;
            else
                return true;
        }
        return false;
    }
};

219

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.

使用map

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

220

Given an array of integers, find out whether there are two distinct indices i and j in the array such that the absolute difference between nums[i] and nums[j] is at most t and the absolute difference between i and j is at most k.

这个题比较操蛋的就是有个数据相减超过了INT_MAX，所以只能用long long过渡一下

class Solution {
public:
    bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) 
    {
        for (int i = 0; i < nums.size(); i++)
        {
            for (int j = i + 1; j < nums.size() && j <= i + k; j++)
            {
                if (my_abs(nums[i], nums[j], t))
                    return true;
            }
        }
        return false;
    }
private:
    bool my_abs(int a, int b, int c)
    {
        long long aa = a;
        long long bb = b;
        long long cc = c;
        long long ab = aa - bb;
        if (ab >= 0 && ab <= cc)
            return 1;
        else if (ab < 0 && -ab <= cc)
            return 1;
        return 0;        
    }
};