[LeetCode] Contains Duplicate(II,III)

Contains Duplicate

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.

解题思路

用一个set保存数组中的值，假设发现当前值已经在set中存在。则返回true。

实现代码

// Rumtime: 67 ms
class Solution {
public:
    bool containsDuplicate(vector<int>& nums) {
        set<int> s;
        for (int i = 0; i < nums.size(); i++)
        {
            if (s.insert(nums[i]).second == false)
            {
                return true;
            }
        }

        return false;
    }
};

Contains Duplicate II

Given an array of integers and an integer k, find out whether there 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.

解题思路

用map存储数组元素和下标。看是否存在与当前元素相等且下标之差小于等于k的元素，存在则返回true。否则将当前元素和其下标存入map。

实现代码

// Runtime: 76 ms
class Solution {
public:
    bool containsNearbyDuplicate(vector<int>& nums, int k) {
        map<int, int> mymap;
        for (int i = 0; i < nums.size(); i++)
        {
            if (mymap.find(nums[i]) != mymap.end() && i - mymap[nums[i]] <= k)
            {
                return true;
            }
            else
            {
                mymap[nums[i]] = i;
            }
        }

        return false;
    }
};

Contains Duplicate III

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

解题思路

用一个multiset存储数组中的元素，保持multiset中元素个数为k。这样multiset中元素与当前元素num[i]的下标之差均小于等于k。

iterator lower_bound (const value_type& val) const能够用于返回值小于等于val的元素的迭代器，推断返回迭代器所指向的值与当前元素之差是否小于等于t就可以。

实现代码

// Runtime: 44 ms
class Solution {
public:
    bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) {
        multiset<int> s;
        for (int i = 0; i < nums.size(); i++)
        {
            if (s.size() == k + 1)
            {
                s.erase(s.find(nums[i-k-1]));
            }
            auto it = s.lower_bound(nums[i] - t);
            if (it != s.end())
            {
                int diff = nums[i] > *it ? nums[i] - *it : *it - nums[i];
                if (diff <= t)
                {
                    return true;
                }
            }

            s.insert(nums[i]);
        }

        return false;
    }
};