leetcode majority elements

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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.

解法:

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

When we move the pointer forward over 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 is the current candidate.

When we are done, the current candidate is the majority element, 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

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原文地址：http://www.cnblogs.com/lilyfindjobs/p/4186491.html