LeetCode Majority Element

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.

思路分析：求主元素的经典算法是Moore投票算法，基本可以理解为将主元素和不同元素一一对应相消，最后剩下来的元素就是主元素。具体做法是，初始计数器times和candidate都是0,然后遍历数组，当times为0时置换新的candidate，遇到和candidate相同元素times++，遇到和candidate相异元素times--，如此操作，如果主元素存在，最后剩下的一定是主元素。时间复杂度O（n）,空间复杂度O（1）.

AC Code

public class Solution {
    public int majorityElement(int[] nums) {
        //0627
        //Moore Voting 
        int n = nums.length;
        
        int candidate = 0;
        int times = 0;
        for(int i = 0; i < n; i++){
            if(times == 0) candidate = nums[i];
            if(nums[i] != candidate){
                times--;
            } else{
                times++;
            }
        }
        return candidate;
        //0631
    }
}

LeetCode Majority Element

原文：http://blog.csdn.net/yangliuy/article/details/46968121