image.png

    思路：
    1、 按照位数分段，观察每一段的递归数量关系。
    代码：方法1没有解决越界问题，方法2正确；

    • 注意越界问题，该用double用double
    • +1也会出现越界
    1. public static void main(String[] args) {
    2.         System.out.println("123".charAt(1));
    3.         int a = 123;
    4.         System.out.println(String.valueOf(a).charAt(0));
    5.         System.out.println(findNthDigit(10));;
    6.         System.out.println(findNthDigit(10));;
    7.         System.out.println(findNthDigit(10));;
    8.         System.out.println(findNthDigit(1000));;
    9.         System.out.println(findNthDigit2(1000000000));;
    10.         System.out.println(findNthDigit2(2147483647));;
    11.     }
    12.     public  static int findNthDigit(int n) {
    13.         if(n <= 9) return n;
    14.         int l  =2;
    15.         int upBound = (int)(Math.pow(10 , l ));
    16.         int howMany = (int)(Math.pow(10 , l )) - (int)(Math.pow(10 , l-1 ));
    17.         int howManyDigits = howMany * l ;
    18.         int cur = 10 ;
    19.         /**先定位到几位数*/
    20.         while(cur <= n+1 - howManyDigits){
    21.             cur += howManyDigits;
    22.             l++;
    23.             howMany = (int)(Math.pow(10 , l )) - (int)(Math.pow(10 , l-1 ));
    24.             howManyDigits = howMany * l ;
    25.         }
    26.         int left = n+1 - cur;
    27.         /**定位第几个数字*/
    28.         int quotient =  left/l;
    29.         int remainder = left % l;
    30.         int num = (int)(Math.pow(10,l-1)) + quotient -1;
    31.         if(remainder != 0) {
    32.             num ++;
    33.             return Integer.valueOf(String.valueOf(String.valueOf(num).charAt(remainder-1)));
    34.         }else {
    35.             String numStr = String.valueOf(num);
    36.             char c = numStr.charAt(numStr.length()-1);
    37.             return Integer.valueOf(String.valueOf(c));
    38.         }
    39.     }
    41.     public  static int findNthDigit2(int n) {
    42.         if(n <= 9) return n;
    43.         int l  =2;
    44.         int upBound = (int)(Math.pow(10 , l ));
    45.         double howMany = (int)(Math.pow(10 , l )) - (int)(Math.pow(10 , l-1 ));
    46.         double howManyDigits = howMany * l ;
    47.         double cur = 9 ;
    48.         /**先定位到几位数*/
    49.         while(cur <= n - howManyDigits){
    50.             cur += howManyDigits;
    51.             l++;
    52.             howMany = (Math.pow(10 , l )) - (Math.pow(10 , l-1 ));
    53.             howManyDigits = howMany * l ;
    54.         }
    55.         int left = (int) (n - cur);
    56.         /**定位第几个数字*/
    57.         int quotient =  left/l;
    58.         int remainder = left % l;
    59.         int num = (int)(Math.pow(10,l-1)) + quotient -1;
    60.         if(remainder != 0) {
    61.             num ++;
    62.             return Integer.valueOf(String.valueOf(String.valueOf(num).charAt(remainder-1)));
    63.         }else {
    64.             String numStr = String.valueOf(num);
    65.             char c = numStr.charAt(numStr.length()-1);
    66.             return Integer.valueOf(String.valueOf(c));
    67.         }
    69.     }