标签：position between erase nta abs 分享 dea elements pre
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：
1 class Solution { 2 public: 3 bool containsNearbyDuplicate(vector<int>& nums, int k) { 4 if (nums.empty()) { 5 return false; 6 } 7 8 unordered_map<int, int> my_map; 9 for (int i = 0; i < nums.size(); ++i) { 10 11 if (my_map.find(nums[i]) == my_map.end()) { 12 my_map[nums[i]] = i; 13 } 14 else { 15 if (abs(i - my_map[nums[i]]) <= k) { 16 return true; 17 } 18 } 19 } 20 21 return false; 22 } 23 };
解法一：
1 class Solution { 2 public: 3 bool containsNearbyDuplicate(vector<int>& nums, int k) { 4 if (nums.empty()) { 5 return false; 6 } 7 8 unordered_map<int, int> my_map; 9 for (int i = 0; i < nums.size(); ++i) { 10 if (my_map.find(nums[i]) != my_map.end() && abs(i - my_map[nums[i]]) <= k) { 11 return true; 12 } 13 my_map[nums[i]] = i; 14 } 15 16 return false; 17 } 18 };
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：
1 class Solution { 2 public: 3 bool containsNearbyDuplicate(vector<int>& nums, int k) 4 { 5 unordered_set<int> s; 6 7 if (k <= 0) return false; 8 if (k >= nums.size()) k = nums.size() - 1; 9 10 for (int i = 0; i < nums.size(); i++) 11 { 12 if (i > k) s.erase(nums[i - k - 1]); 13 if (s.find(nums[i]) != s.end()) return true; 14 s.insert(nums[i]); 15 } 16 17 return false; 18 } 19 };
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.
219. Contains Duplicate II【easy】
标签：position between erase nta abs 分享 dea elements pre
原文地址：http://www.cnblogs.com/abc-begin/p/7623453.html
