题解 | #在二叉树中找到两个节点的最近公共祖先#
整数中1出现的次数（从1到n整数中1出现的次数）
http://www.nowcoder.com/practice/bd7f978302044eee894445e244c7eee6
public class Solution{ public int NumberOf1Between1AndN_Solution(int n){ int count = 0,bitNum = 1,high = n /10,cur = n % 10,low = 0; while(cur != 0 || high != 0){ if(cur < 1){ count += high * bitNum; }else if(cur == 1){ count += high*bitNum +low +1; }else{ count += (high+1)*bitNum; } low += cur * bitNum; cur = high % 10; high = high /10; bitNum *= 10; } return count; } }
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