# 2283. Check if Number Has Equal Digit Count and Digit Value

Easy
You are given a 0-indexed string `num` of length `n` consisting of digits.
Return `true` if for every index `i` in the range `0 <= i < n`, the digit `i` occurs `num[i]` times in `num`, otherwise return `false`.
Example 1:
Input: num = "1210"
Output:
true
Explanation:
num = '1'. The digit 0 occurs once in num.
num = '2'. The digit 1 occurs twice in num.
num = '1'. The digit 2 occurs once in num.
num = '0'. The digit 3 occurs zero times in num.
The condition holds true for every index in "1210", so return true.
Example 2:
Input: num = "030"
Output:
false
Explanation:
num = '0'. The digit 0 should occur zero times, but actually occurs twice in num.
num = '3'. The digit 1 should occur three times, but actually occurs zero times in num.
num = '0'. The digit 2 occurs zero times in num.
The indices 0 and 1 both violate the condition, so return false.
Constraints:
• `n == num.length`
• `1 <= n <= 10`
• `num` consists of digits.

### 解題

Runtime: 3 ms, faster than 35.71%
Memory Usage: 1.9 MB, less than 100.00%
func digitCount(num string) bool {
m := make(map[byte]int)
for i:=0; i<len(num); i++ {
m[num[i]]++
}
for i:=0; i<len(num); i++ {
if m[byte(i+48)]!= int(num[i]-48) {
return false
}
}
return true
}