LeetCode(169) Majority Number

题目如下：

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.

分析如下：

居然几年前的一道baidu面试题也出现在leetcode上了，而且其实也来自几年前的《编程之美》中的找水王那到题目。这道题目的思路直接摘抄自官网：

Runtime: O(n) — Moore voting algorithm: We maintain a current candidate and a counter initialized to 0. As we iterate the array, we look at the current element x:
If the counter is 0, we set the current candidate to x and the counter to 1.
If the counter is not 0, we increment or decrement the counter based on whether x is the current candidate.
After one pass, the current candidate is the majority element. Runtime complexity = O(n).

我的代码:

//90ms
class Solution { 
public:
    int majorityElement(vector<int> &num) {
        if (num.size() == 0) return 0;
        int count = 1;
        int majority_number = num[0];
        for (int i = 1; i < num.size(); ++i) {
            if (num[i] == majority_number) {
                count++;
            } else if (count == 0) {
                majority_number = num[i]; 
            }else{
                count--;
            }
        }
        return majority_number;
    }
};

写完之后，发现这道题目还有官方的解答，里面个思路还挺有意思，用到了bit运算，虽然对这道题目而言，不一定合适，但是思路挺有意思。

Runtime: O(n) — Bit manipulation: We would need 32 iterations, each calculating the number of 1's for the ith bit of all n numbers. Since a majority must exist, therefore, either count of 1's > count of 0's or vice versa (but can never be equal). The majority number’s ith bit must be the one bit that has the greater count.