多数元素-LeetCode

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//法一 先快排 然后循环排出
/*int cmp (const void *a, const void *b) {
    return *(int *)a-*(int *)b;
}
int majorityElement(int* nums, int numsSize){
    int cnt=0;
    int majorityElement;
    qsort(nums,numsSize,sizeof(nums[0]),cmp);
    for(int i=0;i<numsSize;i++){
        if(majorityElement!=nums[i]){
            majorityElement=nums[i];
            cnt=0;
        }
        cnt++;
        if(cnt>numsSize/2) break;
        
    }
    return majorityElement;
}*/
//法二 阵地消耗 一对一消耗
int majorityElement(int* nums, int numsSize){
    int cnt=0;
    int majorityElement=nums[0];
    for(int i=0;i<numsSize;i++){
        if(majorityElement==nums[i]){
            cnt++;
        }else{
            cnt--;
        }
        if(cnt<=0){
            majorityElement=nums[i+1];
        }
    }
    return majorityElement;
}

 

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ps:C中qsort( )函数的用法https://www.cnblogs.com/tsingke/p/5347672.html