Observe the pattern, concise java code


  • 0
    D

    My solution is not the best on all posts, just try to share my thinking process

    public class Solution {
        public String originalDigits(String s) {
            String[] dict = new String[]{"zero","one","two","three","four","five","six","seven","eight","nine"};
            int[] index = new int[]{0,6,7,5,2,4,3,1,8,9};
            int[] result = new int[10];
            int[] map = new int[26];
            for(char a: s.toCharArray()) map[a-'a']++;
            StringBuilder res = new StringBuilder();
            for(int i =0;i<10  ;i++){
                int cur = index[i];
                String temp = dict[cur];
                int min= Integer.MAX_VALUE;
                for(char a: temp.toCharArray()) min = Math.min(min,map[a-'a']);
                for(char a: temp.toCharArray()) {map[a-'a'] = map[a-'a']-min;}
                result[cur] = min;
            }
            
            for(int i =0 ;i<result.length;i++) {
                while(result[i]-->0) res.append(String.valueOf(i));
            }
            return res.toString();
        }
    }
    

    Explain:
    I mainly wanna explain index array part
    0:First we start to count how many zero are in the String since character "z" is unique from one to ten
    6:same apply to six where character "x" is unique
    7:after exclude 6 ,we find that char "s" in seven is unique
    5: after exclude 7, we find that char "v" is unique in five
    2: char "w" in unique in two, you can put 2 in the front, does not matter
    4: after exclude five, char "f" is unique in four
    3: after exclude two, we find char "t" is unique in three
    1: "o" is unique in one
    8: "h" is unique in eight
    then 9

    rest of the code, I think is pretty straightforward

    for index array, there are other combination, you may try yourself


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