Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You may assume no duplicates in the array.

Example 1:

Input: [1,3,5,6], 5
Output: 2
Example 2:

Input: [1,3,5,6], 2
Output: 1
Example 3:

Input: [1,3,5,6], 7
Output: 4
Example 4:

Input: [1,3,5,6], 0
Output: 0

int searchInsert(int* nums, int numsSize, int target){
    for(int i = 0; i < numsSize; i++) {
        if(*(nums+i) == target)
            return i;
        else if(*(nums + i) > target)
            return i;
        else {
            continue;
        }
    }
   return numsSize;
}

The count-and-say sequence is the sequence of integers with the first five terms as following:

1.     1
2.     11
3.     21
4.     1211
5.     111221
1 is read off as "one 1" or 11.
11 is read off as "two 1s" or 21.
21 is read off as "one 2, then one 1" or 1211.

Given an integer n where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence.

Note: Each term of the sequence of integers will be represented as a string.

 

Example 1:

Input: 1
Output: "1"
Example 2:

Input: 4
Output: "1211"

var countAndSay = function(n) {
  let res = "1"
  let bPostion = ''
  let count = 0
  let temp = []
  for (let i = 2; i <= n ; i++ ){
    rt = res + '!'
    for(let j of rt) {
      if ((bPostion || j) == j) {
        count++;
      }
      else {
        temp.push(String(count || 1) + (bPostion || j))
        count = 1
      }
      bPostion = j
    }
    if(temp.length) {
      res = ''
      for (let i of temp){
        res += i
      }
    }
    bPostion = ''
    temp = []
    count = 0
  }

  return res;
};

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output: 321
Example 2:

Input: -123
Output: -321
Example 3:

Input: 120
Output: 21
Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231,  231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

#include <limits.h>
int reverse(int x){
    long long temp, res = 0;
    while(x) {
        temp = x % 10;
        x /= 10;
        res = (res * 10) + temp;
    }
    return (res < INT_MIN || res > INT_MAX) ? 0 : res;
}