Majority Element

方法一:排序取中间的那个数就行O(n*log(n));

int cmp ( const void *a , const void *b){
        return *(int *)a - *(int *)b;
}
int majorityElement(int num[], int n) {
    qsort(num,n,sizeof(num[0]),cmp);
    return num[n/2];
}

方法二:编程之美上好像有这个解法

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