1. Title
Given an integer n, count the number of occurrences of the number 1 in all non-negative integers less than or equal to n.
Example:
输入: 13
输出: 6
解释: 数字 1 出现在以下数字中: 1, 10, 11, 12, 13 。
Two, solve
1. Brute force cracking
Ideas:
After reading the topic, you can write the code directly, but the method times out .
Code:
class Solution {
public int countDigitOne(int n) {
int cnt = 0;
for (int i = 1; i <= n; i++) {
String str = Integer.toString(i);
for (char c : str.toCharArray()) {
if (c == '1') cnt++;
}
}
return cnt;
}
}
Time complexity: O (nlogn) O(nlogn)O ( n l o g n )
space complexity: O (logn) O(logn)O ( l o g n )
2. Dynamic planning
version 1
Ideas:
Main reference 3, the specific analysis is as follows:
The number Y is the number of X digits, each of the digits (one, ten, hundred, thousand, ten thousand...) is composed of nxnx − 1… n 2 n 1 n_xn_{x−1}…n_2n_1nxnx−1⋯n2n1. among them
- 1 number : cnt;
- Current position : ni n_ini, Marked as cur;
- High bit : nxnx − 1… ni + 1 n_xn_{x−1}…n_{i+1}nxnx−1…ni+1, Marked as high;
- Low bits : ni − 1… n 2 n 1 n_{i−1}…n_2n_1ni−1…n2n1, Marked as low;
- Bit factor : 1 0 i 10^i10i , marked as factor.
1. if cur = 0，then cnt = high×digit
eg Find the tens of 2304, that is, when facor=10, the number of 1 occurrences.
另一种推导：十位固定位1
1) 千位选{
0，1}，百位{
0，1，...，9}，个位{
0，1，...，9}，cnt = 2*10*10=200;
2) 千位选2，百位{
0，1，2}，个位{
0，1，...，9}，cnt = 1*3*10=30;
总计：200+30=230. 可推出公式：cnt = high×digit
2. if cur = 1，then cnt = high×digit+low+1
eg Find the tens of 2314, that is, when facor=10, the number of 1 occurrences.
另一种推导：十位固定位1
1) 千位选{
0，1}，百位{
0，1，...，9}，个位{
0，1，...，9}，cnt = 2*10*10=200;
2) 千位选2， 百位{
0，1，2}， 个位{
0，1，...，9}，cnt = 1*3*10=30;
3）千位选2， 百位选3， 个位{
0，1，...，4}，cnt = 5;
总计：200+30+5=235. 可推出公式：cnt = high×digit+low+1
3. if cur > 1，then cnt = (high+1)×digit
eg Find the tens of 2324, that is, when facor=10, the number of 1 occurrences.
另一种推导：十位固定位1
1) 千位选{
0，1}，百位{
0，1，...，9}，个位{
0，1，...，9}，cnt = 2*10*10=200;
2) 千位选2， 百位{
0，1，2，3}， 个位{
0，1，...，9}，cnt = 1*4*10=40;
总计：200+40=240. 可推出公式：cnt = (high+1)×digit
Code:
class Solution {
public int countDigitOne(int n) {
int res = 0;
long a = 0;
long b = 0;
for(long m=1;m<=n;m*=10){
a = n/m;
b = n%m;
if(a % 10 > 1){
res += a/10 * m + m;
}else if( a%10 == 1){
res += a/10 * m + b + 1;
}else{
res += a/10 * m;
}
}
return res;
}
}
Time complexity: O (logn) O(logn)O ( l o g n )
space complexity: O (1) O(1)O ( 1 )
Version 2
Idea: To further optimize the above derivation formula.
Code:
// V1.0
class Solution {
public int countDigitOne(int n) {
if (n <= 0) return 0;
long ones = 0;
for (long i = 1, q = n; i <= n; i *= 10, q /= 10) {
long pre = n / (i * 10), cur = q % 10, suf = n % i;
ones += pre * i;
ones += (1 < cur ? i : (1 == cur ? suf + 1: 0));
}
return (int) ones;
}
}
// V2.0
class Solution {
public int countDigitOne(int n) {
if (n <= 0) return 0;
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;
}
}
Time complexity: O (logn) O(logn)O ( l o g n )
space complexity: O (1) O(1)O ( 1 )
Three, reference
1. The number of number
1 2. 4+ lines, O(log n), C++/Java/Python
3. Interview question 43. The number of 1 to n integers (clear illustration)
4. Runtime: 0 ms, faster than 100.00% of Java online
5. Comparison of the efficiency of int to String and String to int methods in Java