(6/1/16)Leetcode 229. Majority Element II

Medium, 用时20分钟, 还是Boyer-Moore Majority Vote algorithm.

1. 此题可扩展为k的情况
2. error1, count1应设置为1
3. error2, 未考虑corner case长度为1, 从而p1 = p2, list中有两个元素
4. 本题思路比较巧妙, 构思时考虑corner case：
   [2,2,9,9,1,3,4,5,6,7,8,2,2,2,2,2,2,9,9,9,9,9,9,9].
   开始是p1,p2分别为2和9, 经过1,3,4,5,6,7,8后, p1,p2再次回到2和9。p1或p2其中一个移动时，另一个不动，从而保证出现次数大于n/3的存在于p1,p2中。

public class Solution {
    public List majorityElement(int[] nums) {
        List result = new ArrayList<>();
        if(nums == null || nums.length == 0) return result;
        int p1 = nums[0], p2 = nums[0];
        int count1 = 0, count2 = 0;
        int length = nums.length;
        for(int i = 0; i < length; i++){
            if(p1 == nums[i]){
                count1 ++;
            }
            else if(p2 == nums[i]){
                count2 ++;
            }
            //error2: count1 = 1, count2 = 1;
            else if(count1 == 0){
                p1 = nums[i];
                count1 = 1;
            }
            else if(count2 == 0){
                p2 = nums[i];
                count2 = 1;
            }
            else{
                count1 --;
                count2 --;
            }
        }
        count1 = 0;
        count2 = 0;
        for (int i = 0; i< length; i++){
            if(nums[i] == p1) count1++;
            if(nums[i] == p2) count2++;
        }
        if(count1 > length / 3) result.add(p1);
        //error1:这里要加p1 != p2
        if(count2 > length / 3 && p1 != p2) result.add(p2);
        return result;
    }
}

拓展: 用O(n)与O(1)找出array中出现频率最高的数: