[LeetCode]Majority Element

最新推荐文章于 2019-02-22 16:24:01 发布

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本文介绍了一种高效的算法——摩尔投票算法(Moore voting algorithm)，用于找出数组中出现次数超过一半的元素。该算法仅需遍历一次数组并使用两个变量进行维护，即可在O(n)的时间复杂度内找到多数元素。

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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++;
				else
					counter--;
			}
		}

		return candidate;
}