剑指offer_整数中1出现的次数

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题目：

求1~n的数字中，数字1出现的次数。

最无脑的做法，从1遍历到n，累计每个数字中1出现的次数。。。。
在leetcode上看到最牛逼的做法
先上代码再说：

public int countDigitOne(int n) {    int ones = 0;    for (long m = 1; m <= n; m *= 10)        ones += (n/m + 8) / 10 * m + (n/m % 10 == 1 ? n%m + 1 : 0);    return ones;}

就是这么简洁！！!

时间复杂度达到了O（lgn）！！！

看一下leetcode上面大神的解释：

https://discuss.leetcode.com/topic/18054/4-lines-o-log-n-c-java-python

For each position, split the decimal representation into two parts, for example split n=3141592 into a=31415 and b=92 when we’re at m=100 for analyzing the hundreds-digit. And then we know that the hundreds-digit of n is 1 for prefixes “” to “3141”, i.e., 3142 times. Each of those times is a streak, though. Because it’s the hundreds-digit, each streak is 100 long. So (a / 10 + 1) * 100 times, the hundreds-digit is 1.

Consider the thousands-digit, i.e., when m=1000. Then a=3141 and b=592. The thousands-digit is 1 for prefixes “” to “314”, so 315 times. And each time is a streak of 1000 numbers. However, since the thousands-digit is a 1, the very last streak isn’t 1000 numbers but only 593 numbers, for the suffixes “000” to “592”. So (a / 10 * 1000) + (b + 1) times, the thousands-digit is 1.

The case distincton between the current digit/position being 0, 1 and >=2 can easily be done in one expression. With (a + 8) / 10 you get the number of full streaks, and a % 10 == 1 tells you whether to add a partial streak.

根本在于利用数学对数字的分析，直接推出了简洁的数学公式。

在苏宁的笔试中，遇到过这么一道题，n个人中有m对好友（两两称为一对），朋友的朋友都算是同一个朋友圈。问有多少个朋友圈。

咋一看，毫无头绪，其实也是道数学题。
当m>n时，只有1个朋友圈；
当m<n时，有n-m个朋友圈
就是这么简单。。。。。。。。。。。。。

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