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.

#include<stdio.h>

void sort(int num[], int low, int high){
    if(low < high){
        int val = num[low], i = low, j = high;
        while(i < j){
            while(num[j] >= val && i < j) j--;
            if(i < j) num[i++] = num[j]; 
            while(num[i] < val && i < j) i++;
            if(i < j) num[j--] = num[i];
        }
        num[i] = val;
        sort(num, low, i-1);
        sort(num, i+1, high);
    }
}
//先排序再找，时间复杂度太高，超时了
int majorityElement1(int num[], int n) {
    sort(num, 0, n-1);
    return num[n/2];
}

//这个方法accepted了

int majorityElement(int num[], int n) {
    int result = 0;
    int time = 0 ,i;
    for(i = 0; i < n; i++) {
        if(time == 0) {
            result = num[i];
            time++;
        }
        else {
            if(num[i] == result) time++;
            else time--;
        }
    }
    return result;
}

void main(){
    int num[] = {3,3,3,3,3,3,3,1,1,1,22,122,11,15,1,2,3,3,3,3,4};
    int n = 21, i;
    printf("%d\n", majorityElement(num,n));
}

Majority Element