[LeetCode 229] Majority element II

標籤：leetcode

Given an integer array of size n, find all elements that appear more than? n/3 ? times. The algorithm should run in linear time and in O(1) space.

Solution:

At most has two elements in the result, can use two counter to record occurrence. Then calculate if match the condition.

public List<Integer> majorityElement(int[] nums) {        List<Integer> result = new ArrayList<>();        if(nums.length==0) return result;        int n1 = nums[0];        int n2 = 0;        int count1 = 1;        int count2 = 0;        for(int i=1;i<nums.length;i++){            if(nums[i] == n1){                count1++;            }else if(nums[i] == n2){                count2++;            }else{                if(count1 == 0){                    n1 = nums[i];                    count1++;                }else if(count2 == 0){                    n2 = nums[i];                    count2++;                }else{                count1--;                count2--;            }            }        }        count1 = 0;        count2 = 0;        for(int i=0;i<nums.length;i++){            if(nums[i] == n1) count1++;            if(nums[i] == n2) count2++;        }        if(count1>nums.length/3) result.add(n1);        if(n1 == n2) return result;        if(count2>nums.length/3) result.add(n2);        return result;    }

著作權聲明：本文為博主原創文章，未經博主允許不得轉載。

[LeetCode 229] Majority element II