【leetcode】Majority Element (easy)(*^__^*)

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:

找主要元素,用major记录主要字母,n记录major相对于其他字母多出现的次数

if         n == 0 , 设当前数字为主要数字

else if   当前数字等于主要数字, n++

else     当前数字不等于主要数字, n--

因为主要数字出现多于一半,所以最后major表示的一定是主要数字。

没测试就一次AC了 

class Solution {
public:
    int majorityElement(vector<int> &num) {
        int major;
        int n = 0;

        for(int i = 0; i < num.size(); i++)
        {
            if(n == 0)
            {
                major = num[i]; n++;
            }
            else if(major == num[i])
            {
                n++;
            }
            else
            {
                n--;
            }
        }
        return major;
    }
};
原文地址:https://www.cnblogs.com/dplearning/p/4214245.html