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LeetCode

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Problem

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.

Code

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