14. There is a ten-digit number. Counting from left to right, if its first digit is a certain number, then this ten-digit number contains that many 0s; if its second digit is a certain number, then this ten-digit number contains that many 1s; if its third digit is a certain number, then this ten-digit number contains that many 2s; and so on; if its tenth digit is a certain number, then this ten-digit number contains that many 9s. This ten-digit number is $\qquad$
【Analysis】Consider the tenth digit of this number. If it is 1, then this number contains one 9. Consider the position of 9. If it is not in the second position, then this ten-digit number, besides 1 and 9, has 9 other digits, totaling 11 numbers, which is impossible; if it is in the second position, then this ten-digit number should have nine 1s, so except for the second position, the remaining numbers are all 1s. Since the first position is 1, there should be 1 zero, which would exceed 11 numbers, which is impossible;
If the tenth digit is 2 or more, then this number contains 2 or more 9s, which would exceed 10 numbers, which is impossible.
Therefore, the tenth digit of this number is 0.
Similarly, the ninth and eighth digits of this number are also 0.
Consider the sum of the digits of this ten-digit number, which should be 10 (the sum of these 10 digits is the sum of the occurrences of 0, 1, 2, ..., 9 in this ten-digit number).
Now we know that the eighth, ninth, and tenth digits are 0.
Consider the first digit, which is at least 3.
If the first digit is 3, then there are six remaining digits, with a sum of 7, and there cannot be any 0s. Therefore, these six digits can only be $1+1+1+1+1+2=7$, with 5 ones. The second digit should be 5, but there is no 5 among these six digits, which is clearly impossible, so the first digit is not 3.
If the first digit is 4, then there are six remaining digits, with a sum of 6, and there is one 0. Therefore, these six digits can only be $0+1+1+1+1+2=6$, with 4 ones. The second digit should be 4, but there is no 4 among these six digits, which is clearly impossible, so the first digit is not 4.
If the first digit is 5, then there are six remaining digits, with a sum of 5, and there are two 0s. Therefore, these six digits can only be $0+0+1+1+1+2=5$, with 3 ones. The second digit should be 3, but there is no 3 among these six digits, which is clearly impossible, so the first digit is not 4.
If the first digit is 6, then there are six remaining digits, with a sum of 4, and there are three 0s. Therefore, these six digits can only be $0+0+0+1+1+2=4$, with 2 ones. The second digit should be 2, and there is a 2 among these six digits, so the second digit is 2. Since there is 1 two, the third digit is 1. Since the first digit is 6, the seventh digit is 1. The fourth, fifth, and sixth digits are 0. After verification, 6210001000 meets all the requirements.
Since the eighth, ninth, and tenth digits are all 0, this ten-digit number does not contain 7, 8, or 9, so the first digit is at most 6. In summary, the required ten-digit number is unique, which is 6210001000.