leetcode-2283-easy

Check if Number Has Equal Digit Count and Digit Value

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.

思路一：map 存储，然后遍历，没有难度，就是题目稍微有点绕

public boolean digitCount(String num) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < num.length(); i++) {
        map.compute(num.charAt(i) - '0', (k, v) -> v == null ? 1 : v + 1);
    }

    for (int i = 0; i < num.length(); i++) {
        if (map.getOrDefault(i, 0) != num.charAt(i) - '0') {
            return false;
        }
    }

    return true;
}