Leetcode majority elements

Given an array of size n, find the majority element. The majority element is the element, the appears more than times ⌊ n/2 ⌋ .

Assume that the array was non-empty and the majority element always exist in the array.

Solution:

As we sweep we maintain a pair consisting of a current candidate and a counter. Initially, the current candidate are unknown and the counter is 0.

When we move the pointer forward to an element e:

• If The counter is 0, we set the current candidate to e and we set the counter to 1.
• If The counter is not 0, we increment or decrement the counter according to whether e are the current candidate.

When we do, the current candidate is the majority element, and if there is a majority.

public class Solution {public    int majorityelement (int[] num) {    int sum=0;    int result=0;    for (int i=0;i<num.length;i++) {    if (sum==0) {    result=num[i];    sum++;    }    else if (Result==num[i]) {    sum++;    if (SUM>NUM.LENGTH/2) {    return result;    }    }    else sum--;    }    return result;}    }

　　

Leetcode majority elements