219. Contains Duplicate II【easy】

219. Contains Duplicate II【easy】

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.

错误解法：

 class Solution {
 public:
     bool containsNearbyDuplicate(vector<int>& nums, int k) {
         if (nums.empty()) {
             return false;
         }

         unordered_map<int, int> my_map;
         for (int i = ; i < nums.size(); ++i) {

             if (my_map.find(nums[i]) == my_map.end()) {
                 my_map[nums[i]] = i;
             }
             else {
                 if (abs(i - my_map[nums[i]]) <= k) {
                     return true;
                 }
             }
         }

         return false;
     }
 };

解法一：

 class Solution {
 public:
     bool containsNearbyDuplicate(vector<int>& nums, int k) {
         if (nums.empty()) {
             return false;
         }

         unordered_map<int, int> my_map;
         for (int i = ; i < nums.size(); ++i) {
             if (my_map.find(nums[i]) != my_map.end() && abs(i - my_map[nums[i]]) <= k) {
                 return true;
             }
             my_map[nums[i]] = i;
         }

         return false;
     }
 };

参考@jianchao.li.fighter 的代码

Well, the basic idea is fairly straightforward. We maintain a mapping mp from a value in nums to its position (index) i. Each time we meet an unseen value, we add it to the map (mp[nums[i]] = i). Otherwise, depending on whether the recorded index mp[nums[i]] and the current index i satisfy i - mp[nums[i]] <= k (node that the new index i is larger than the old index mp[nums[i]]), we return true or update the index (mp[nums[i]] = i). If all the elements have been visited and we have not returned true, we will return false.

解法二：

 class Solution {
 public:
     bool containsNearbyDuplicate(vector<int>& nums, int k)
     {
        unordered_set<int> s;

        if (k <= ) return false;
        if (k >= nums.size()) k = nums.size() - ;

        for (int i = ; i < nums.size(); i++)
        {
            if (i > k) s.erase(nums[i - k - ]);
            if (s.find(nums[i]) != s.end()) return true;
            s.insert(nums[i]);
        }

        return false;
     }
 };

参考@luo_seu 的代码，就是维持固定那么多长度的数值

The basic idea is to maintain a set s which contain unique values from nums[i - k] to nums[i - 1],
if nums[i] is in set s then return true else update the set.