Question 1: duplicate element

Problem 2: The power of two

Question 3: Move the zero

Question 4: missing numbers

Question 5: The first wrong version

Question 6: masseur

Question 7: 3 power

Question 8: seeking 1 + 2 + ... + n

Item 9: 0 value becomes the number of steps

Question 10: How much less than the current number of digits

Stay button (LeetCode) regularly brush title, each lasting 10 questions, heavy traffic comrades can look at ideas I share, is not the most efficient solution, but only to enhance each other.

Question 1: duplicate element

Questions requirements are as follows:

Answer (C language):

int compare(const void *a, const void *b){
    return *(int*)a - *(int*)b;
}

bool containsDuplicate(int* nums, int numsSize){
    qsort(nums, numsSize, sizeof(int), compare);
    for(int i = 0; i < numsSize - 1; i++){
        if(nums[i] == nums[i+1]){
            return true;
        }
    }
    return false;
}

Operating efficiency as follows:

Problem 2: The power of two

Questions requirements are as follows:

Answer (C language):

bool isPowerOfTwo(int n){
    while(n%2==0&&n>1){
        n=n/2;
    }

    if(n==1){
        return true;
    }
    else{
        return false;
    }
}

Operating efficiency as follows:

Question 3: Move the zero

Questions requirements are as follows:

Answer (C language):

void moveZeroes(int* nums, int numsSize){

    for(int i=0,j = 0,temp=0; j < numsSize;++j) {
        if(nums[j] != 0) {
            temp = nums[i];
            nums[i] = nums[j];
            nums[j] = temp;
            ++i;
        }
    }
}

Operating efficiency as follows:

Question 4: missing numbers

Questions requirements are as follows:

Answer (C language):

int missingNumber(int* nums, int numsSize){
    int ans = numsSize*(numsSize + 1)/2, i;

    for(i = 0; i < numsSize; i++){
        ans -= nums[i];
    }
    return ans;
}

Operating efficiency as follows:

Question 5: The first wrong version

Questions requirements are as follows:

Answer (C language):

// Forward declaration of isBadVersion API.
bool isBadVersion(int version);

int firstBadVersion(int n) {
    long low=1;
    long high=n;
    
    while(low<high){
        long mid=(low+high)/2;
        if(!isBadVersion(mid)){
            low=mid+1;
        }
        else{
            high=mid;
        }
    }
    return low;
}

Operating efficiency as follows:

Question 6: masseur

Questions requirements are as follows:

Answer (C language):

//设dp[i]表示第i个房屋之前的最高金额，则求dp[numsSize-1]即可。
//状态转移方程：

//dp[0] = nums[0]
//dp[1] = max(nums[1], nums[0])
//dp[i] = max(dp[i-2] + nums[i], dp[i-1])(i>=2)

int massage(int* nums, int numsSize){
    if(!nums || numsSize == 0)
        return 0;

    if(numsSize == 1)
        return nums[0];

    int *dp = (int*)malloc(sizeof(int) * numsSize);
    dp[0] = nums[0];
    dp[1] = fmax(nums[1], nums[0]);

    for(int i = 2; i < numsSize; i++){
        dp[i] = fmax(dp[i-2] + nums[i], dp[i-1]);
    }
    
    return dp[numsSize - 1];
}

Operating efficiency as follows:

Question 7: 3 power

Questions requirements are as follows:

Answer (C language):

bool isPowerOfThree(int n){
    if (n == 0) 
        return false;

    while(n%3==0){
        n/=3;
    }

    if(n==1)
        return true;
    
    return false;
}

Operating efficiency as follows:

Question 8: seeking 1 + 2 + ... + n

Questions requirements are as follows:

Answer (C language):

//（1）如果多个变量均非0（包括None、False等），那么返回最后一个变量的值。如3 and 2 and 'a'的返回值为'a'；
//（2）如果多个变量中存在0值，则返回第一个0值。如1 and 'a' and 0 and None的返回值为0。


int sumNums(int n){
    int res = n;
    n && (res += sumNums(n-1));
    return res;
}

Operating efficiency as follows:

Item 9: 0 value becomes the number of steps

Questions requirements are as follows:

Answer (C language):

int numberOfSteps (int num){
    int cou=0;
    
    while(num!=0){
        if(num%2==0){
            num/=2;
        }
        else{
            num-=1;
        }
        cou++;
    }

    return cou;
}

Operating efficiency as follows:

Question 10: How much less than the current number of digits

Questions requirements are as follows:

Answer (C language):

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* smallerNumbersThanCurrent(int* nums, int numsSize, int* returnSize){
    int* data_buf=(int*)malloc(sizeof(int)*(numsSize+1));
    memset(data_buf,'\0',sizeof(int)*(numsSize+1));

    for(int i=0;i<numsSize;i++){
        for(int j=0;j<numsSize;j++){
            if(nums[j]<nums[i])
                data_buf[i]++;
        }
    }

    data_buf[numsSize]='\0';
    *returnSize=numsSize;

    return data_buf;
}

Operating efficiency as follows: