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. Hint: Beware of overflow. c…

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. 解题思路: 递归 static public in…

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. 解法:参考编程之美 132页 2.4 1的数目,以下代…

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. 链接: http://leetcode.com…

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. 代码: class Solution {…

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. 按不同位置统计 31456 统计百位时: (0-31)…

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. 方法一 class Solution { public:…

这题属于需要找规律的题.先想一下最简单的情形:N = 10^n - 1 记X[i]表示从1到10^i - 1中 1 的个数,则有如下递推公式:X[i] = 10 * X[i - 1] + 10^(i - 1) 这个递推公式可以这么观察得到: i = 0, X[0] = 0 i = 1, 从1到9, X[1] = 1 i = 2, 从1到99, X[2] = 20:可以设想,把所有数都写成两位数(比如1写成01, 2写成02),我们暂且不统计最高位的1, 则首先至少有10 * X[1]个1,然后我…

给定一个整数 n,计算所有小于等于 n 的非负数中数字1出现的个数. 例如: 给定 n = 13, 返回 6,因为数字1出现在下数中出现:1,10,11,12,13. 详见:https://leetcode.com/problems/number-of-digit-one/description/ Java实现: 方法一: class Solution { public int countDigitOne(int n) { StringBuilder sb=new StringBuilder()…

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. Hint: Beware of overflow.…