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Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋
times.
You may assume that the array is non-empty and the majority element always exist in the array.
public class Solution { public int majorityElement(int[] nums) { int[] dup = new int[nums.length]; int sum = 0; int min = 100000; int index = 0; for(int i = 0;i < nums.length;i++){ for(int k = 0;k < nums.length;k++){ dup[k] = nums[k]; } for(int j = 0;j < nums.length;j++){ dup[j] -= nums[i]; sum += Math.abs(dup[j]); } System.out.println("sum:-->"+sum); if(sum <= min){ min = sum; index = i; } sum = 0;//用于本轮计数的注意清零 } return nums[index]; }}