思路
给出leetcode的思路,说的更为详细
https://leetcode-cn.com/problems/1nzheng-shu-zhong-1chu-xian-de-ci-shu-lcof/solution/mian-shi-ti-43-1n-zheng-shu-zhong-1-chu-xian-de-2/
public class Solution { public int NumberOf1Between1AndN_Solution(int n) { int digit = 1, res = 0; int high = n / 10, cur = n % 10, low = 0; while(high != 0 || cur != 0) { if(cur == 0) res += high * digit; else if(cur == 1) res += high * digit + low + 1; else res += (high + 1) * digit; low += cur * digit; cur = high % 10; high /= 10; digit *= 10; } return res; } }