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[0] = '1'. The digit 0 occurs once in num. num[1] = '2'. The digit 1 occurs twice in num. num[2] = '1'. The digit 2 occurs once in num. num[3] = '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] = '0'. The digit 0 should occur zero times, but actually occurs twice in num. num[1] = '3'. The digit 1 should occur three times, but actually occurs zero times in num. num[2] = '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.

• 把字符串中的每個數(shù)字都放到map中，然后遍歷map中的數(shù)字跟值去判斷是否是一樣的，如果不一樣則返回flag。

• 簡單的題目，但是通過的空間跟時間復(fù)雜度都是不好的。。

Runtime:?13 ms, faster than?6.62%?of?Java?online submissions for?Check if Number Has Equal Digit Count and Digit Value.

Memory Usage:?42.7 MB, less than?6.74%?of?Java?online submissions for?Check if Number Has Equal Digit Count and Digit Value.