440.cpp

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sample 4 ms submission
class Solution {
public:
    int numsToSkip(long currNum, int n){
      long nextNum=currNum+1;  
      int total = 0;  
      while (currNum<=n){
        if (nextNum>n){
           total+=(n-currNum+1);
           break; 
        } else {
          total+=(nextNum-currNum);
          currNum*=10;
          nextNum*=10;  
        }    
      }   
      return total;  
    }
    
    int findKthNumber(int n, int k) {
      int currNum=1;
      int numsPast = 0;  
      while (numsPast<(k-1)){
        int skip = numsToSkip((long)currNum,n);
        if (skip+numsPast<k) {
          currNum++;
          numsPast+=skip;  
        }
        else {
          currNum*=10; 
          numsPast++;  
        }
      }
      return currNum;  
    }
};
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sample 8340 kb submission
class Solution {
public:
    int findKthNumber(int n, int k) {
        int curr = 1;
        while (k > 1) {
            int count = countNums(curr, n);
            if (k > count) { // 走右兄弟节点
                curr++;
                k -= count;
            } else { // 不是当前节点，走最左子节点
                curr *= 10;
                k--;
            }
        }
        return curr;
    }
    
    // 以数p为根的子树含多少[1,n]间的数？
    // 可累加[p,p+1)、[p *10,(p+1) *10)、[p *100,(p+1) *100)、... 与[1,n]重合的部分
    int countNums(int p, long n) {
        int count = 0;
        long start = p, end = p + 1;
        while (start <= n) {
            count += min(end, n + 1) - start;
            start *= 10, end *= 10;
        }
        return count;
    }
};
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