标签：
Count the number of k‘s between 0 and n. k can be 0 - 9.
if n=12, in [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], we have FIVE 1‘s (1, 10, 11, 12)
Analysis:
Method 1: directly count every number.
Method 2: Analytical solution.
Every time, calculate how many k has appears on a specific postion, i.e., on 1, 10, 100,.....
Solution 1:
1 class Solution { 2 /* 3 * param k : As description. 4 * param n : As description. 5 * return: An integer denote the count of digit k in 1..n 6 */ 7 public int digitCounts(int k, int n) { 8 int[] record = new int[10]; 9 Arrays.fill(record,0); 10 for (int i=0;i<=n;i++){ 11 String temp = Integer.toString(i); 12 for (int j=0;j<temp.length();j++){ 13 int ind = (int) (temp.charAt(j)-‘0‘); 14 record[ind]++; 15 } 16 } 17 return record[k]; 18 19 } 20 };
Solution 2:
1 class Solution { 2 /* 3 * param k : As description. 4 * param n : As description. 5 * return: An integer denote the count of digit k in 1..n 6 */ 7 public int digitCounts(int k, int n) { 8 int res = 0; 9 int base = 1; 10 while (base<=n){ 11 int part1 = n/(base*10); 12 if (base>1 && k==0 && part1>0) part1--; 13 part1 *= base; 14 int bar = n/base%10; 15 int part2 = 0; 16 if (k<bar) part2 = base; 17 else if (k==bar) part2 = n%base+1; 18 if (k==0 && n<base*10) part2 = 0; 19 res += part1+part2; 20 base*=10; 21 } 22 return res; 23 } 24 };
标签：
原文地址：http://www.cnblogs.com/lishiblog/p/4194962.html
