leetcode_169_Majority Element

版权所有，欢迎转载，转载请注明出处，谢谢

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.

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

//vs2012测试代码#include<iostream>#include<vector>//方法二：如果不随机会超时；同样思路，随机挑选一个元素，检查是否是多数元素。时间复杂度：Average：O(n)。期望查找次数 <2using namespace std;#define n 5class Solution {public:    int majorityElement(vector<int> &num) {while(1){int randnum;int count=0;//1个元素时，num.size()-1为0。而rand()%0不合法，应该规避。    if(num.size()==1)        return num[0];randnum = rand() % (num.size()-1);for(int i=0; i<num.size(); i++){if( num[i] == num[randnum])count++;if(count > num.size()/2){cout<<num[i];return num[i];}}}    }};int main(){int a;vector<int> num;for(int i=0; i<n; i++){cin>>a;num.push_back(a);}Solution lin;lin.majorityElement(num);}

//方法一：每找出两个不同的element，就成对删除即count--，最终剩下的一定就是所求的。时间复杂度：O(n)class Solution {public:    int majorityElement(vector<int> &num) {int element;        int length=num.size();int count=0;for(int i=0; i<length; i++){if(count == 0){element=num[i];count=1;}else{if(num[i] == element)count++;elsecount--;}}cout<<element;return element;    }};

//方法二：随机挑选一个元素，检查是否是多数元素。时间复杂度：Average：O(n)。期望查找次数 <2class Solution {public:    int majorityElement(vector<int> &num) {    while(1){int randnum;int count=0;//1个元素时，num.size()-1为0。而rand()%0不合法，应该规避。    if(num.size()==1)        return num[0];randnum = rand() % (num.size()-1);for(int i=0; i<num.size(); i++){if( num[i] == num[randnum])count++;if(count > num.size()/2){cout<<num[i];return num[i];}}}    }};

//方法：不随机，超时class Solution {public:    int majorityElement(vector<int> &num) {        int length=num.size();vector<int> count;count.resize(length);count.assign(length,0);for(int i=0; i<length; i++){for(int j=0; j<length; j++){if(num[j] == num[i])count[i]++;}if(count[i] > length/2){cout<<num[i];return num[i];}}    }};