Majority Element

题目链接https://leetcode.com/problems/majority-element/

 

这道题最一开始的思路是建立一个hashmap，这样就是 O(n) time和 O(n) space。

但是后来看了一下leetcode给出的官方解答，发现moore voting algorithm很有意思，就记了下来，以后遇到这个问题就用他这个方法了。

 

以下摘自leetcode

“

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:

1. If the counter is 0, we set the current candidate to x and the counter to 1.

　　2. 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).

”

博主的code

class Solution {

public:

    int majorityElement(vector<int> &num) {

        int candidate = num[0];

        int counter = 0;

        for(int i = 0; i < num.size(); i++) {

            if(counter == 0) {

                candidate = num[i];

                counter = 1;

            }

            if(candidate == num[i])

                counter++;

            else

                counter--;

        }

        return candidate;

    }

};

至于这个算法为什么是正确的，证明在发明者的这篇paper中 http://www.cs.rug.nl/~wim/pub/whh348.pdf