169. Majority Element

public class Solution {
    public int majorityElement(int[] nums) {
        Arrays.sort(nums);
        int len=nums.length;
        int result=0;
        for(int i=0;i<len/2+1;i++)
        if(nums[i]==nums[len/2+i])
            {result=nums[i];
            break;}
         return result;
        
    }
   
}
/*
There are several things needed to pay attention to.
1.upper index and bottom index
if length is an even number;
for example 6; it needs more than 3 same elements, which is four
the idex of the  last element you need to check is 2; which is length/2-1
if length is an odd number;
for example 5;it needs more than 2 same elements, which is 3;
the index of the last element you need to check is 2 ,which is length/2;
Thus if you want to put it together, the index limit should be less than len/2+1;
Further more;
if I write like this;
if(nums[i]==nums[len/2+i])
          
         return nums[i];
         it will say missing return statement;
         
         Sometimes if it says illegal start of the type in the return line; basically you misplaced the brace
*/

更简单的做法：
public class Solution {
    public int majorityElement(int[] nums) {
        Arrays.sort(nums);
        return nums[nums.length / 2];
    }
}
因为如果length是偶数，nums.length/2前面有nums.length/2个元素，就算从0开始都是一样的，也会包括nums.length/2
如果length是奇数，nums.length/2前面和后面有相同的数目的元素，都是（length-1）/2;
所以无论怎样，都会渠道nums.length/2