Question

Source

• leetcode: Number of Digit One | LeetCode OJ
• lintcode: (3) Digit Counts

Count the number of k's between 0 and n. k can be 0 - 9.

Example
if n=12, k=1 in [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12],
we have FIVE 1's (1, 10, 11, 12)

题解

leetcode 上的有点简单，这里以 Lintcode 上的为例进行说明。找出从 0 至整数 n 中出现数位 k 的个数，与整数有关的题大家可能比较容易想到求模求余等方法，但其实很多与整数有关的题使用字符串的解法更为便利。将整数 i 分解为字符串，然后遍历之，自增 k 出现的次数即可。

C++

class Solution {
public:
  /*
   * param k : As description.
   * param n : As description.
   * return: How many k's between 0 and n.
   */
  int digitCounts(int k, int n) {
    char c = k + '0';
    int count = 0;
    for (int i = k; i <= n; i++) {
      for (auto s : to_string(i)) {
        if (s == c) count++;
      }
    }
    return count;
  }
};

Java

class Solution {
  /*
   * param k : As description.
   * param n : As description.
   * return: An integer denote the count of digit k in 1..n
   */
  public int digitCounts(int k, int n) {
    int count = 0;
    char kChar = (char)(k + '0');
    for (int i = k; i <= n; i++) {
      char[] iChars = Integer.toString(i).toCharArray();
      for (char iChar : iChars) {
        if (kChar == iChar) count++;
      }
    }

    return count;
  }
}

源码分析

太简单了，略

复杂度分析

时间复杂度 O(n×L)O(n \times L)O(n×L), L 为 n 的最大长度，拆成字符数组，空间复杂度 O(L)O(L)O(L).