[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.

数组求过半数的元素。

做到这道题的时候，心里也是一阵痛，因为上学期算法期末考刚好有这样一道题，而且有空间的限制。

当时想着用map来实现，时间上是快啊。但是不用想，那么大的数据，空间肯定会爆。

LeetCode上的这个解法只用维护两个变量，而且时间也是O（n）：

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).

源代码：

int majorityElement(vector<int> &num) {int counter=0, candidate;for (int i = 0; i < num.size(); i++) {if (counter == 0) {candidate = num[i];counter++;}else {if (num[i] == candidate)counter++;elsecounter--;}}return candidate;}