标签:10 digit 数字 level int number numbers 400
在无限的整数序列 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...中找到第 n
位数字。
注意:n
是正数且在 32 位整数范围内(n < 231
)。
示例 1:
输入:3 输出:3
示例 2:
输入:11 输出:0 解释:第 11 位数字在序列 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... 里是 0 ,它是 10 的一部分。
Java
class Solution { public int findNthDigit(int n) { int level=1;//The level of numbers, we start from one. long numbers=9;//The number of current digital levels, we start with a number, there are 9 numbers (1-9), it is a single number //Number [1-9] (with 9 digits) is a one-digit number, and the number [10-99] (with 90 digits) is a two-digit number. //The number [100-999] (with 900 numbers) is three digits, first we should find what level //(I mean which number (one digit, two digits, etc.) the nth digit), once we find the number level, //We have already implemented half of the process. If the n-digit * count > 0, it means that the nth bit is not at the current digital level. //We should increase the number level to pass more numbers while(n-level*numbers>0){ //Every time we pass numbers at the current number level n-=level*numbers; level++; // The count grows as follows, 9, 90, 900, 9000 ..... because the count may overflow, so I use the long type numbers*=10; } // After the loop, n represents the nth number of the current baseNumber //The base is 1, 10, 100, 1000, 10000, etc. int baseNumber=(int)Math.pow(10,level-1); // find the number where the nth digit is located int number=(n-1)/level+baseNumber; // Find the number with the nth digit above it int mod=(n-1)%level; return String.valueOf(number).charAt(mod)-'0'; } }
cpp
class Solution { public: int findNthDigit(int n) { int level=1; long numbers=9; while(n-level*numbers>0){ n-=level*numbers; level++; numbers*=10; } int baseNumber=pow(10,level-1); int number=(n-1)/level+baseNumber; int mod=(n-1)%level; return std::to_string(number)[mod]-'0'; } };
py
class Solution: def findNthDigit(self, n: int) -> int: level=1 numbers=9 while n-level*numbers>0: n-=level*numbers level+=1 numbers*=10 baseNumber=10**(level-1) number=(n-1)/level+baseNumber mod=(n-1)%level return str(number)[mod]
标签:10,digit,数字,level,int,number,numbers,400 来源: https://www.cnblogs.com/xxxsans/p/14416449.html
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