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数据结构 -- 顺序表(数组)题目

目录

数组的相关面试题

点击进入>> 移除元素(LeetCode)

代码示例(C语言)

int removeElement(int* nums, int numsSize, int val){
    int idx = 0;

        for(int i = 0; i < numsSize ;++i)
        {
            if(nums[i] != val)
            {
                nums[idx++] = nums[i];
            }
        }
        return idx;
}

点击进入>> 删除有序数组中的重复项(LeetCode)

代码示例(C语言)

int removeDuplicates(int* nums, int numsSize){
    if(numsSize<=1)
         return numsSize;

    int idx = 1;
    for(int i = 1; i< numsSize; ++i)
    {
        if(nums[i] == nums[i-1])
            continue;
        else
        {
            nums[idx++] = nums[i];
        }

    }
    
    return idx;
}

点击进入>> 合并两个有序数组(LeetCode)

代码示例(C语言)

void merge(int* nums1, int nums1Size, int m, int* nums2, int nums2Size, int n){
   int i = m - 1, j = n - 1, idx = nums1Size - 1;
   
   while(i>=0 && j>=0)
   {
       if(nums1[i]<nums2[j])
       {
           nums1[idx--] = nums2[j--];
       }
       else
       {
           nums1[idx--] = nums1[i--];
       }
   }
   if(j>=0)
   memcpy(nums1,nums2,sizeof(int)*(j+1));

点击进入>> 轮转数组(LeetCode)

代码示例(C语言)

void rotate(int* nums, int numsSize, int k){
    if(k >= 0)
    {
        k = ((numsSize>=k)?k:(k%numsSize));
        reverse(nums, 0, numsSize-k-1);
        reverse(nums, numsSize-k, numsSize-1);
        reverse(nums, 0, numsSize-1);

    } 

}
void reverse(int* nums, int begin, int end)
{
    while(begin<end)
    {
        nums[begin] = nums[begin]^nums[end];
        nums[end] = nums[begin]^nums[end];
        nums[begin++] = nums[begin]^nums[end--];
    }
}

点击进入>> 数组形式的整数加法(LeetCode)

代码示例(C语言)

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* addToArrayForm(int* num, int numSize, int k, int* returnSize){
    //计算整数k的位数
    int len =0, kk = k;
    while(kk != 0)
    {
        len++;
        kk /= 10;
    }
    //申请数组
    int arrlen = (numSize>len)?(numSize+1):(len+1);
    int* ret = (int*)malloc(sizeof(int)*arrlen);

    //进位
    int step = 0;
    int end = numSize -1;
    *returnSize = 0;

    while(arrlen-1 -(*returnSize) != 0 )
    {
        //逐位相加
        if(end>=0 && (num[end]+k%10 + step)>=10)
        {
            ret[arrlen-1 -(*returnSize) ] = (num[end]+k%10 + step) % 10;
            step = 1;
        }
        else if(end>=0 && (num[end]+k%10 + step)<10)
        {
            ret[arrlen-1- (*returnSize)] = (num[end]+k%10) + step;
            step = 0;
        }
        else
        {
            if((k%10 + step)>=10)
        {
            ret[arrlen-1 -(*returnSize) ] = (k%10 + step) % 10;
            step = 1;
        }
        else if((k%10 + step)<10)
        {
            ret[arrlen-1- (*returnSize)] = (k%10) + step;
            step = 0;
        }
        }

        if(k>0)
        k/=10;
        if(end>=0)
        end--;

        (*returnSize)++;
    }
   if(step == 1)
   {
       ret[0] = 1;
       (*returnSize)++;
   }

    if((*returnSize) == arrlen)
    return ret;
    else
    {
         int* ret1 = (int*)malloc(sizeof(int)*(*returnSize));
         memcpy(ret1, ret+1,sizeof(int)*(*returnSize) );
         return ret1;
    }



}
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