In this Leetcode Contains Duplicate II problem solution we have given an integer array nums and an integer k, return true if there are two distinct indices i and j in the array such that nums[i] == nums[j] and abs(i - j) <= k.
Problem solution in Python.
class Solution: def containsDuplicate(self, nums): """ :type nums: List[int] :rtype: bool """ return len(list(set(nums))) < len(nums)
Problem solution in Java.
class Solution { public boolean containsNearbyDuplicate(int[] nums, int k) { int end = 0; for (int i = 0; i < nums.length - 1; i++) { end = i + 1; while (end < nums.length && nums[end] != nums[i]) { end++; } //no duplicate value for nums[i] if (end >= nums.length) continue; if (nums[end] == nums[i] && end - i <= k) { return true; } } return false; } }
Problem solution in C++.
class Solution { public: bool containsNearbyDuplicate(vector<int>& nums, int k) { unordered_set<int> set; for(int i=0; i<nums.size(); i++) { if(set.count(nums[i])>0) return true; set.insert(nums[i]); if(set.size()>k) { set.erase(nums[i-k]); } } return false; } };
Problem solution in C.
typedef struct Node{ int latest_index; int key; int val; struct Node *next; }; #define SIZE 1000 int getPos(int n){ long long num=n; if(num<0) num=-1*num; return num%SIZE; } bool containsNearbyDuplicate(int* nums, int numsSize, int k){ struct Node** hashMap=calloc(SIZE+1, sizeof(struct Node *)); for(int i=0;i<numsSize;i++){ int pos=getPos(nums[i]); struct Node* iter=hashMap[pos]; struct Node* prev=hashMap[pos]; while(iter){ if(iter->key==nums[i]){ if(iter->val>=1){ if(i <= iter->latest_index+k){ return true; } } iter->latest_index=i; iter->val++; break; } prev=iter; iter=iter->next; } if(iter==NULL){ struct Node *tmp=malloc(sizeof(struct Node)); tmp->key=nums[i]; tmp->val=1; tmp->latest_index=i; tmp->next=NULL; if(prev){ prev->next=iter; } else{ hashMap[pos]=tmp; } } } return false; }
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