LeetCode 刷题第一天

1Given 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.
2
3You may assume no duplicates in the array.
4
5Example 1:
6
7Input: [1,3,5,6], 5
8Output: 2
9Example 2:
10
11Input: [1,3,5,6], 2
12Output: 1
13Example 3:
14
15Input: [1,3,5,6], 7
16Output: 4
17Example 4:
18
19Input: [1,3,5,6], 0
20Output: 0

1int searchInsert(int* nums, int numsSize, int target){
2    for(int i = 0; i < numsSize; i++) {
3        if(*(nums+i) == target)
4            return i;
5        else if(*(nums + i) > target)
6            return i;
7        else {
8            continue;
9        }
10    }
11   return numsSize;
12}

1The count-and-say sequence is the sequence of integers with the first five terms as following:
2
31.     1
42.     11
53.     21
64.     1211
75.     111221
81 is read off as "one 1" or 11.
911 is read off as "two 1s" or 21.
1021 is read off as "one 2, then one 1" or 1211.
11
12Given an integer n where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence.
13
14Note: Each term of the sequence of integers will be represented as a string.
15
16 
17
18Example 1:
19
20Input: 1
21Output: "1"
22Example 2:
23
24Input: 4
25Output: "1211"

1var countAndSay = function(n) {
2  let res = "1"
3  let bPostion = ''
4  let count = 0
5  let temp = []
6  for (let i = 2; i <= n ; i++ ){
7    rt = res + '!'
8    for(let j of rt) {
9      if ((bPostion || j) == j) {
10        count++;
11      }
12      else {
13        temp.push(String(count || 1) + (bPostion || j))
14        count = 1
15      }
16      bPostion = j
17    }
18    if(temp.length) {
19      res = ''
20      for (let i of temp){
21        res += i
22      }
23    }
24    bPostion = ''
25    temp = []
26    count = 0
27  }
28
29  return res;
30};

1Given a 32-bit signed integer, reverse digits of an integer.
2
3Example 1:
4
5Input: 123
6Output: 321
7Example 2:
8
9Input: -123
10Output: -321
11Example 3:
12
13Input: 120
14Output: 21
15Note:
16Assume 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.

1#include <limits.h>
2int reverse(int x){
3    long long temp, res = 0;
4    while(x) {
5        temp = x % 10;
6        x /= 10;
7        res = (res * 10) + temp;
8    }
9    return (res < INT_MIN || res > INT_MAX) ? 0 : res;
10}