LeeCode-Majority Element

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.

int majorityElement(int* nums, int numsSize) 
{
    if(numsSize==1)
        return nums[0];
    
    
    int i,j,k;
    for(i=1;i<numsSize;i++)
    {
        for(j=i-1;j>=0;j--)
        {
            if(nums[i]>=nums[j])
                break;
        }

        if(i!=j-1)
        {
            int temp=nums[i];
            for(k=i-1;k>j;k--)
            {
                nums[k+1]=nums[k];
            }
            nums[k+1]=temp;
        }
    }

    int Time;
    Time=numsSize/2;
    int count=1;
    for(i=0;i<numsSize-1;i++)
    {
        if(nums[i]==nums[i+1])
        {
            count++;
            if(count>Time)
            {
                return nums[i];
            }
        }
        else
        {
            count=1;
        }
    }
    

    return 0;
}
LeeCode-Majority Element_第1张图片

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