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.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

public class Solution {
    public int majorityElement(int[] nums) {
        return helper(nums, 0, nums.length-1);
    }
//确定返回值类型(根据在下一次递归中的作用, 或者题目的返回值类型)
private int helper(int[] nums, int lo, int hi) {
//递归出口
        if (lo == hi) return nums[lo];
        int mid = lo + (hi - lo) / 2;
//分治
        int left = helper(nums, lo, mid);
        int right = helper(nums, mid+1, hi);
//开始对分治后的小分支进行操作: 根据题目要求用TC写一下, 准备进行返回了
// 优化的好啊, 如果相等不用判断了
        if (left == right) return left;
        else {
            int l = 0, r = 0;
            for (int i = lo; i <= hi; i++) {
                if (nums[i] == left)  l++;
                if (nums[i] == right) r++;
            }
            return l > r ? left : right;
        }
    }
}

public int majorityElement(int[] nums) {
        int majority = nums[0], count = 1;
        for (int i = 1; i < nums.length; i++) {
            if (count == 0) {
                majority = nums[i];
                count++;
            } else if (nums[i] == majority) {
                count++;
            } else {
                count--;
            }
        }
        return majority;
}

public class Solution {
    public int majorityElement(int[] num) {
        int n = num.length;
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        for (int elem : num) {
            if (map.containsKey(elem)) {
                map.put(elem, map.get(elem)+1);
            }
            else {
                map.put(elem, 1);
            }
        }
        for (int item : map.keySet()) {
            if (map.get(item) > n/2) {
                return item;
            }
        }
        return -1;
    }
}