LeetCode OJ 之 Number of Digit One (数字1的个数)
2017-05-25 21:20
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题目:
Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.For example:
Given n = 13,
Return 6, because digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.
思路:
对这个数字的每一位求存在1的数字的个数。从个位開始到最高位。
举个样例54215。比方如今求百位上的1,54215的百位上是2。能够看到xx100到xx199的百位上都是1,这里xx从0到54。即100->199, 1100->1199...54100->54199, 这些数的百位都是1,因此百位上的1总数是55*100
假设n是54125,这时因为它的百位是1,先看xx100到xx199。当中xx是0到53,即54*100, 然后看54100到54125,这是26个。所以百位上的1的总数是54*100 + 26.
假设n是54025。那么仅仅须要看xx100到xx199中百位上的1,这里xx从0到53,总数为54*100
求其它位的1的个数的方法是一样的。
代码:
class Solution { public: int countDigitOne(int n) { int res=0; long left, right, base=1; if (n <= 0) return 0; while (n >= base) { left = n / base; //left包括当前位 right = n % base; //right为当前位的右半边 if ((left % 10) > 1) res+= (left / 10 + 1) * base; else if ((left % 10) == 1) res+= (left / 10) * base+ (right + 1); else res+= (left / 10) * base; base *= 10; } return res; } };能够把上面三个条件合成一步。例如以下:
class Solution { public: int countDigitOne(int n) { int res=0; long left, right, base=1; if (n<=0) return 0; while (n>=base) { left = n / base; //left包括当前位 right = n % base; //right为当前位的右半边 res += ((left + 8) / 10 * base) + (left % 10 == 1) * (right + 1); base *= 10; } return res; } };
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