Problem 1: Search Insert location

Problem 2: Appearance array

Question 3: and maximum subsequence

Question 4: The length of the last word

Question 5: plus one

Question 6: Binary sum

Question 7: square root

Question 8: climbing stairs

Question 9: remove sorts the list of repeating elements

Question 10: Merge two ordered arrays

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.

Development: stay button (LeetCode) brush questions, simple questions (the first phase)

Problem 1: Search Insert location

Questions requirements are as follows:

Answer (C language):

int searchInsert(int* nums, int numsSize, int target){
    int num=0;

    if(numsSize<=0)
        return 0;

    for(int i=0,j=1;i<numsSize;i++,j++){
        if(target<nums[0]){
            num=0;
            break;
        }
        else if(target>nums[numsSize-1]){
            num=numsSize;
            break;
        }
        else{
            if(nums[i]==target) {
                num=i; 
                break;
            }
            if(target>nums[i] && target<nums[j]){
                num=j; 
                break;
            } 
        }
    }

    return num;
}

Problem 2: the appearance of an array

Questions requirements are as follows:

Answer (C language):

char * countAndSay(int n){
    
    if(n==1) 
        return "1";
    
    char *s=countAndSay(n-1);

    char *a=(char *)malloc(5000);
    char *ans=(char *)malloc(5000);

    int count=0;
    int l=strlen(s);
    int j=0;
    for(int i=0;i<l;i++){
        if(i == 0 || s[i]==s[i-1])
            count++;
        else{
            a[j++]='0'+count;
            a[j++]=s[i-1];
            count=1;
        }
        if (i == l - 1) {
            a[j++]='0'+count;
            a[j++]=s[i];
        }
    }
    
    a[j]='\0';
    ans = a;
    return ans;
}

Question 3: The maximum subsequence and

Questions requirements are as follows:

Answer (C language):

int maxSubArray(int* nums, int numsSize){
  int max = nums[0];
  int b = 0;
  for(int i = 0; i < numsSize; i++){
    b += nums[i];
    if(b > max) max = b;
    if(b < 0) b = 0;
  }
  return max;
}

Question 4: The length of the last word

Questions requirements are as follows:

Answer (C language):

int lengthOfLastWord(char * s){  
    int num=strlen(s); 
    int len=0;

    if(num<=0)
        return 0;

    //重后往前，若有空格，则去掉空格
    //若全为空格则返回0
    while(s[--num]==' '){
        if(num==0){
            return 0;
        }
    }

    while(num>=0){
        if(s[num--]==' ')
            break;
        len++;
    }   

    return len;
}

Question 5: plus one

Questions requirements are as follows:

Answer (C language):

int* plusOne(int* digits, int digitsSize, int* returnSize){
    
    int *buf=(int *)malloc((digitsSize + 1) * sizeof (int));

    for(int i=digitsSize-1;i>=0;i--){
        if(digits[i]+1==10){
            digits[i]=0;
        }
        else{
            digits[i]+=1;
            returnSize[0]=digitsSize;
            return digits;
        }
    }
    
    buf[0]=1;
    for(int i=0,j=1;i<digitsSize;i++,j++)
        buf[j]=digits[i];

    returnSize[0]=digitsSize+1;
    return buf;
}

Question 6: Binary sum

Questions requirements are as follows:

Answer (C language):

char * addBinary(char * a, char * b){
    int num=0,len_a=strlen(a),len_b=strlen(b),len_long=0;

    if(len_a>=len_b){
        len_long=len_a+2;
    }
    else{
        len_long=len_b+2;
    }

    char *buf=(char *)malloc(sizeof(char)*(len_long)) ;
    buf[--len_long]='\0';
    buf[0]='0';

    while(len_long>0){
        num+=len_a>0?a[--len_a]-'0':0;
        num+=len_b>0?b[--len_b]-'0':0;

        if(num==0){
            num=0;
            buf[--len_long]='0';
        }
        else if(num==1){
            num=0;
            buf[--len_long]='1';
        }
        else if(num==2){
            num=1;
            buf[--len_long]='0';
        }
        else if(num==3){
            num=1;
            buf[--len_long]='1';
        }
    }
    
    if(buf[0]=='0')
        return buf+1;
    else    
        return buf;
}

Question 7: square root

Questions requirements are as follows:

Answer (C language):

int mySqrt(int x){
    //牛顿迭代法
    long r = x;
	while(r * r > x)
	{
		r = (r + x / r) / 2;//迭代
	}
	return (int)r;
}

Question 8: climbing stairs

Questions requirements are as follows:

Answer (C language), algorithm description:

int climbStairs(int n){
    if (n == 1) {
        return 1;
    }
    int *dp=(int *)malloc(sizeof(int)*(n+1));
    dp[1] = 1;
    dp[2] = 2;

    for (int i = 3; i <= n; i++) {
        dp[i] = dp[i - 1] + dp[i - 2];
    }
    
    return dp[n];
}

Question 9: remove sorts the list of repeating elements

Questions requirements are as follows:

Answer (C language):

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

struct ListNode* deleteDuplicates(struct ListNode* head){
    struct ListNode * p = head;
    struct ListNode * q;
    
    while(NULL != p)
    {
        while(NULL != p->next && p->val == p->next->val)
        {
            q = p->next;
            p->next = q->next;
            free(q);
        }
        p = p->next;
    }
    return head;
}

Question 10: Merge two ordered arrays

Questions requirements are as follows:

Answer (C language):

void merge(int* nums1, int nums1Size, int m, int* nums2, int nums2Size, int n){
    int p = m + n;
    m--;
    n--;

    while (n >= 0 && m >= 0) {
        nums1[--p] = nums1[m] <= nums2[n] ? nums2[n--] : nums1[m--];
    }
    
    while (n >= 0) {
        nums1[--p] = nums2[n--];
    }
}