2283. 判断一个数的数字计数是否等于数位的值

class Solution:
    def digitCount(self, num: str) -> bool:
        cnt = Counter(num)
        return all(cnt[str(i)] == int(v) for i, v in enumerate(num))

class Solution {
    public boolean digitCount(String num) {
        int[] cnt = new int[10];
        int n = num.length();
        for (int i = 0; i < n; ++i) {
            ++cnt[num.charAt(i) - '0'];
        }
        for (int i = 0; i < n; ++i) {
            if (cnt[i] != num.charAt(i) - '0') {
                return false;
            }
        }
        return true;
    }
}

class Solution {
public:
    bool digitCount(string num) {
        int cnt[10]{};
        for (char& c : num) {
            ++cnt[c - '0'];
        }
        for (int i = 0; i < num.size(); ++i) {
            if (cnt[i] != num[i] - '0') {
                return false;
            }
        }
        return true;
    }
};

func digitCount(num string) bool {
	cnt := [10]int{}
	for _, c := range num {
		cnt[c-'0']++
	}
	for i, v := range num {
		if cnt[i] != int(v-'0') {
			return false
		}
	}
	return true
}

function digitCount(num: string): boolean {
    const n = num.length;
    const count = new Array(10).fill(0);
    for (let i = 0; i < n; i++) {
        count[i] = Number(num[i]);
    }
    for (const c of num) {
        count[c]--;
    }
    return count.every(v => v === 0);
}

impl Solution {
    pub fn digit_count(num: String) -> bool {
        let s = num.as_bytes();
        let n = num.len();
        let mut count = [0; 10];
        for i in 0..n {
            count[i] = s[i] - b'0';
        }
        for c in s {
            count[(c - b'0') as usize] -= 1;
        }
        count.iter().all(|v| *v == 0)
    }
}

bool digitCount(char* num) {
    int count[10] = {0};
    for (int i = 0; num[i]; i++) {
        count[i] = num[i] - '0';
    }
    for (int i = 0; num[i]; i++) {
        count[num[i] - '0']--;
    }
    for (int i = 0; i < 10; i++) {
        if (count[i] != 0) {
            return false;
        }
    }
    return true;
}