Nth Digit(leetcode 400)

Find the nth digit of the infinite integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...

Note:
n is positive and will fit within the range of a 32-bit signed integer (n < 231).

Example 1:

Input:
3

Output:
3

Example 2:

Input:
11

Output:
0

Explanation:
The 11th digit of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... is a 0, which is part of the number 10.

 

// Time Limit Exceeded  我想一個個加上去老是能夠的吧  可是效率不行
public static  int findNthDigit(int n) {

    int num = 0;
    for(int i=1;i<=n;i++){
        num += (i+"").length();
        if(num >= n){
            return (i+"").charAt((i+"").length()-num+n-1)-'0';
        }
    }
    return 0;

}

 

//換一個方式 找到那個我要取的數
public int findNthDigit2(int n) {
    int len = 1;
    long count = 9;
    int start = 1;

    while (n > len * count) {
        n -= len * count;
        len += 1;
        count *= 10;
        start *= 10;
    }

    start += (n - 1) / len;
    String s = Integer.toString(start);
    return Character.getNumericValue(s.charAt((n - 1) % len));
}

git：https://github.com/woshiyexinjie/leetcode-xin