229. Majority Element II

    /*
     * 229. Majority Element II 
     * 12.7 by Mingyang 
     * 观察可知，数组中至多可能会有2个出现次数超过 ⌊ n/3 ⌋
     * 的众数记变量n1, n2为候选众数； c1, c2为它们对应的出现次数遍历数组，记当前数字为num若num与n1或n2相同，
     * 则将其对应的出现次数加1否则，若c1或c2为0，则将其置为1，对应的候选众数置为num否则，
     * 将c1与c2分别减1最后，再统计一次候选众数在数组中出现的次数，若满足要求，则返回之。
     */
    public List<Integer> majorityElement2(int[] nums) {
        if (nums == null || nums.length == 0)
            return new ArrayList<Integer>();
        List<Integer> result = new ArrayList<Integer>();
        int number1 = nums[0], number2 = nums[0], count1 = 0, count2 = 0, len = nums.length;
        for (int i = 0; i < len; i++) {
            if (nums[i] == number1)
                count1++;
            else if (nums[i] == number2)
                count2++;
            else if (count1 == 0) {
                number1 = nums[i];
                count1 = 1;
            } else if (count2 == 0) {
                number2 = nums[i];
                count2 = 1;
            } else {
 //左边右边同时减减哦！！！！！！！
//这个是我自己做的时候没有料到的！！！！！！！！！！！！！！
                count1--;
                count2--;
            }
        }
        count1 = 0;
        count2 = 0;
        for (int i = 0; i < len; i++) {
            if (nums[i] == number1)
                count1++;
            else if (nums[i] == number2)
                count2++;
        }
        if (count1 > len / 3)
            result.add(number1);
        if (count2 > len / 3)
            result.add(number2);
        return result;
    }