169. Majority Element

Given an array of size n, find the majority element. The majority element is the element, the appears more than times ⌊ n/2 ⌋ .

Assume that the array was non-empty and the majority element always exist in the array.

Solution 1:

Compare naive's ideas, iterate through the array, use the map to save the number and the corresponding number. Then traverse keyset to read out the number greater than N/2.

Map is useful for handling the number of repetitions, and for recording the same key with different value changes.

Attention

int Change=res.get (nums[i]) +1;
Res.put (Nums[i],change);

If the direct res.put (Nums[i],res.get (nums[i]) + +) error occurs. I do not know the specific reason, the individual think is unable to read value and also change the value.

 Public classSolution { Public intMajorityelement (int[] nums) {Map<Integer,Integer> res=NewHashmap<integer,integer>(); intMajority=0;  for(inti=0;i<nums.length;i++)        {            if(!Res.containskey (Nums[i])) {Res.put (nums[i],1); }            Else            {                intChange=res.get (Nums[i]) +1;            Res.put (Nums[i],change); }        }         for(intNum:res.keySet ()) {            if(Res.get (num) > (NUMS.LENGTH/2) ) {Majority=num; }        }        returnmajority; }}

See the solution of discussion, feel oneself too naive = =

Solution 2:

 Public intMajorityElement2 (int[] nums) {Map<integer, integer> MyMap =NewHashmap<integer, integer>(); //Hashtable<integer, integer> myMap = new Hashtable<integer, integer> ();    intRet=0;  for(intnum:nums) {        if(!Mymap.containskey (num)) mymap.put (num,1); Elsemymap.put (num, mymap.get (num)+1); if(Mymap.get (num) >nums.length/2) {ret=num;  Break; }    }    returnret;}

No need to use iterator to check the keyset, the number of records and the number of times can be compared with the size of N/2.

Solution3:

 Public int majorityElement1 (int[] nums) {    arrays.sort (nums);     return NUMS[NUMS.LENGTH/2];}

This Nima is simply cheating!!

Solution4:

Boyer-moore Majority vote algorithm

Https://en.wikipedia.org/wiki/Boyer%E2%80%93Moore_majority_vote_algorithm

The core idea is to save a counter corresponding to canadiate, if there is really a choice times greater than half, then when the end of the cycle counter necessarily corresponding majority choice.

 public  int  MajorityElement3 (int  int  count=0, ret = 0< Span style= "color: #000000;"    >;  for  (int   Num:nums) { if  (Count==0 = num;  if  (Num!=ret) Count --;  else   count  ++;  return   ret;}  

Because the topic has already said certainly exists, therefore does not have to examine is more than half, exits the circulation counter certainly corresponds to the majority.

169. Majority Element