441. Arranging Coins
You have a total of n coins that you want to form in a staircase shape, where every k-th row must have exactly k coins.
Given n, find the total number of full staircase rows that can be formed.
n is a non-negative integer and fits within the range of a 32-bit signed integer.
Example 1:
n = 5
The coins can form the following rows:
¤
¤ ¤
¤ ¤
Because the 3rd row is incomplete, we return 2.
Example 2:
n = 8
The coins can form the following rows:
¤
¤ ¤
¤ ¤ ¤
¤ ¤
Because the 4th row is incomplete, we return 3.
Solution in C++：
关键点：
- 数列求和公式
思路：
- 题目很明显就是一个求和公式的逆运用，这里我主要出的问题可能是因为数据类型写的有点问题，然后sqrt的结果居然是nan，思索好久不懂情况，然后自己拆分之后OK，但是看到别人的写法和我的类似但是AC，所以可能还是自己对于类型转换这样的问题掌握的还不够。
int arrangeCoins(int n) {
// x * (x + 1) / 2 = n => x = -1/2 + sqrt(2*n+1/4)
return sqrt(2.0 * double(n) + 0.25) - 0.5;
}
443. String Compression
Given an array of characters, compress it in-place.
The length after compression must always be smaller than or equal to the original array.
Every element of the array should be a character (not int) of length 1.
After you are done modifying the input array in-place, return the new length of the array.
Follow up:
Could you solve it using only O(1) extra space?
Example 1:
Input:
["a","a","b","b","c","c","c"]
Output:
Return 6, and the first 6 characters of the input array should be: ["a","2","b","2","c","3"]
Explanation:
"aa" is replaced by "a2". "bb" is replaced by "b2". "ccc" is replaced by "c3".
Example 2:
Input:
["a"]
Output:
Return 1, and the first 1 characters of the input array should be: ["a"]
Explanation:
Nothing is replaced.
Example 3:
Input:
["a","b","b","b","b","b","b","b","b","b","b","b","b"]
Output:
Return 4, and the first 4 characters of the input array should be: ["a","b","1","2"].
Explanation:
Since the character "a" does not repeat, it is not compressed. "bbbbbbbbbbbb" is replaced by "b12".
Notice each digit has it's own entry in the array.
Note:
- All characters have an ASCII value in
[35, 126]. 1 <= len(chars) <= 1000.
Solution in C++：
关键点：
- 扫描
思路：
- 思路很简单就是扫描，然后计数，添加到result字符串中，这里当然写也可以通过变量当指针的作用使用。主要需要注意的就是控制变量的合法性。
int compress(vector<char>& chars) {
size_t size = chars.size();
if (size == 0)
return 0;
string result(1, chars[0]);
for(int i = 1; i < size; ++i){
int tmp = 1;
while(chars[i] == chars[i-1] && i < size){
++i;
++tmp;
}
if (tmp > 1)
result += to_string(tmp);
if (i < size)
result += chars[i];
}
for(int i = 0; i < result.size(); ++i)
chars[i] = result[i];
return result.size();
}
小结
今天其实两题应该做的都可以挺快的都是因为自己的一些细节处理的不好，不仅导致心情不好而且做的效率也不高。主要第一题的问题是关于数据类型处理的问题，然后第二题是由于粗心没有控制变量i的合法性，导致一些奇怪现象出现，让自己摸不着头脑。
知识点
- 数据类型转换
- i合法性控制