169.Majority Element （法1排序法2多数投票）

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Given an array of size n, find the majority element. Themajority element is the element that appears more than ? n/2 ? times.

You may assume that the array is non-empty and the majority element alwaysexist in the array.

Credits:
Special thanks to @ts for adding this problem and creating all testcases.

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 Divide and Conquer Array Bit Manipulation

#pragma once
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

//法1：排序，返回中间数
int majorityElement1(vector<int> &num)
{
	sort(num.begin(), num.end());
	return num[num.size() / 2];
}

//法2：多数投票，每找到一对不同的数，就成对删除。
int majorityElement2(vector<int> &num) 
{
	int count = 1;
	int element = num[0];//题设num非空
	for (int i = 1; i < num.size(); i++)
	{
		if (count == 0)
		{
			element = num[i];
			count = 1;
			continue;
		}
		if (num[i] == element)
			count++;
		else
			count--;
	}
	return element;
}
void main()
{
	vector<int> v = { 1,1,1,1,2,2,2 };
	//vector<int>::iterator begain, end;

	/*sort(v.begin(), v.end());
	for (int i = 0; i < v.size(); i++)
	cout << v[i] << ' ';*/
	cout << majorityElement2(v) << endl;
	system("pause");
}

169.Majority Element （法1排序法2多数投票）

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原文地址：http://blog.csdn.net/hgqqtql/article/details/43415899