本问题对应的 leetcode 原文链接：剑指 Offer 43. 1～n 整数中 1 出现的次数
问题描述
输入一个整数 n ，求1～n这n个整数的十进制表示中1出现的次数。
例如，输入12，1～12这些整数中包含1 的数字有1、10、11和12，1一共出现了5次。
示例 1:
输入：n = 12
输出：5
示例 2：
输入：n = 13
输出：6
限制：
1 <= n < 2^31
解题思路
视频讲解直达： 本题视频讲解
代码实现
class Solution {
public int countDigitOne(int n) {
// 几个变量计算：cur = (n/bit)%10, low = n % bit, high = n / bit / 10
// 几个公示
// cur > 1 => (high + 1) * bit
// cur == 1 => (high * bit) + (1 + low)
// cur = 0 => high * bit
long bit = 1;
long sum = 0;
while(bit <= n){
long cur = (n/bit)%10;
long low = n % bit;
long high = n / bit / 10;
if(cur > 1){
sum += (high + 1) * bit;
}else if(cur == 1){
sum += (high * bit) + (1 + low);
}else{
sum += high * bit;
}
bit = bit * 10;
}
return (int)sum;
}
}
时间复杂度：O(logn)
额外空间复杂度：O(1)
Python
class Solution(object):
def countDigitOne(self, n):
"""
:type n: int
:rtype: int
"""
# 几个变量计算：cur = (n/bit)%10, low = n % bit, high = n / bit / 10
# 几个公示
# cur > 1 => (high + 1) * bit
# cur == 1 => (high * bit) + (1 + low)
# cur = 0 => high * bit
bit = 1
sum = 0
while bit <= n:
cur = (n // bit) % 10
low = n % bit
high = n // bit // 10
if cur > 1:
sum += (high + 1) * bit
elif cur == 1:
sum += high * bit + low + 1
else:
sum += high * bit
bit *= 10
return sum
C++
class Solution {
public:
int countDigitOne(int n) {
// 几个变量计算：cur = (n/bit)%10, low = n % bit, high = n / bit / 10
// 几个公式
// cur > 1 => (high + 1) * bit
// cur == 1 => (high * bit) + (1 + low)
// cur = 0 => high * bit
long long bit = 1;
long long sum = 0;
while(bit <= n){
long long cur = (n / bit) % 10;
long long low = n % bit;
long long high = n / bit / 10;
if(cur > 1){
sum += (high + 1) * bit;
}else if(cur == 1){
sum += high * bit + low + 1;
}else{
sum += high * bit;
}
bit *= 10;
}
return sum;
}
};
Go
func countDigitOne(n int) int {
// 几个变量计算：cur = (n/bit)%10, low = n % bit, high = n / bit / 10
// 几个公式
// cur > 1 => (high + 1) * bit
// cur == 1 => (high * bit) + (1 + low)
// cur = 0 => high * bit
var bit, sum int64 = 1, 0
for bit <= int64(n) {
cur := (int64(n) / bit) % 10
low := int64(n) % bit
high := int64(n) / bit / 10
if cur > 1 {
sum += (high + 1) * bit
} else if cur == 1 {
sum += high*bit + low + 1
} else {
sum += high * bit
}
bit *= 10
}
return int(sum)
}
JS
/**
* @param {number} n
* @return {number}
*/
function countDigitOne(n) {
// 几个变量计算：cur = (n/bit)%10, low = n % bit, high = n / bit / 10
// 几个公式
// cur > 1 => (high + 1) * bit
// cur == 1 => (high * bit) + (1 + low)
// cur = 0 => high * bit
let bit = 1;
let sum = 0;
while (bit <= n) {
let cur = Math.floor((n / bit) % 10);
let low = n % bit;
let high = Math.floor(n / bit / 10);
if (cur > 1) {
sum += (high + 1) * bit;
} else if (cur == 1) {
sum += (high * bit) + (1 + low);
} else {
sum += high * bit;
}
bit = bit * 10;
}
return sum;
}
shuaidi