Implement
int sqrt(int x).Compute and return the square root of x, where x is guaranteed to be a non-negative integer.
Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.
Example 1:
Input: 4 Output: 2Example 2:
Input: 8 Output: 2 Explanation: The square root of 8 is 2.82842..., and since the decimal part is truncated, 2 is returned.
整型“开方”
Implement pow(x, n), which calculates x raised to the power n (x^n).
Example 1:
Input: 2.00000, 10 Output: 1024.00000Example 2:
Input: 2.10000, 3 Output: 9.26100Example 3:
Input: 2.00000, -2 Output: 0.25000 Explanation: 2^{-2} = {1/2}^2 = 1/4 = 0.25Note:
- -100.0 < x < 100.0
- n is a 32-bit signed integer, within the range [−2^31, 2^31 − 1]
求x的n次方。
The set
[1,2,3,...,n]contains a total of n! unique permutations.By listing and labeling all of the permutations in order, we get the following sequence for n = 3:
"123""132""213""231""312""321"Given
nandk, return thek-thpermutation sequence.Note:
- Given n will be between 1 and 9 inclusive.
- Given k will be between 1 and n! inclusive.
Example 1:
Input: n = 3, k = 3 Output: "213"Example 2:
Input: n = 4, k = 9 Output: "2314"
假设存在长度为n的数组：[1,2,3,...,n]，输出第k个排列。
Given two non-negative integers
num1andnum2represented as strings, return the product ofnum1andnum2, also represented as a string.Example 1:
Input: num1 = “2”, num2 = “3” Output: “6”
Example 2:
Input: num1 = “123”, num2 = “456” Output: “56088”
将两个字符串表示的数字相乘（大数相乘）。
Divide two integers without using multiplication, division and mod operator.
If it is overflow, return MAX_INT.
不使用乘法、除法、取余实现两个整数相除。
Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.
将罗马数字转换成整型。
Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.
将整型转换成罗马数字形式:
