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 difference between i and j is at most k.
最開始寫了下面這個錯誤代碼,分析了下是對unordered_map理解錯誤。unordered_map是用hash實現的,在內部unordered_map的元素不以鍵值或映射的元素作任何特定的順序排序,其存儲位置取決於哈希值允許直接通過其鍵值爲快速訪問單個元素(具有恆定平均的平均時間複雜度)。可見代碼中 mymap.erase(mymap.begin()); 是不可控的
錯誤方法:
class Solution {
public:
bool containsNearbyDuplicate(vector<int>& nums, int k) {
int n=nums.size();
if(n==0)
return false;
unordered_map<int,int> mymap;
for(int i=0;i<k+1;++i){
mymap[nums[i]]++;
if(mymap[nums[i]]>1)
return true;
}
for(int i=k+1;i<n;++i){
mymap.erase(mymap.begin());
mymap[nums[i]]++;
for(auto i=mymap.begin();i!=mymap.end();i++){
if(i->second > 1)
return true;
}
}
return false;
}
};
方法一:
class Solution {
public:
bool containsNearbyDuplicate(vector<int>& nums, int k)
{
unordered_set<int> s;
if (k <= 0) return false;
if (k >= nums.size()) k = nums.size() - 1;
for (int i = 0; i < nums.size(); i++){
if (i > k) s.erase(nums[i - k - 1]);
if (s.find(nums[i]) != s.end()) return true;
s.insert(nums[i]);
}
return false;
}
};
方法二 :
bool containsNearbyDuplicate(vector<int>& nums, int k) {
unordered_map<int,int> nmap;
for (int i = 0; i <nums.size();i++){
if (nmap.count(nums[i]) == 0)
nmap[nums[i]] = i;
else if (i - nmap[nums[i]] <=k)
return true;
else
nmap[nums[i]] = i;
}
return false;
}