791. Custom Sort String

Problem:
 and T are strings composed of lowercase letters. In S, no letter occurs more than once.
S was sorted in some custom order previously. We want to permute the characters of T so that they match the order that S was sorted. More specifically, if x occurs before y in S, then x should occur before y in the returned string.
Return any permutation of T (as a string) that satisfies this property.
Example :
Input: 
S = "cba"
T = "abcd"
Output: "cbad"
Explanation: 
"a", "b", "c" appear in S, so the order of "a", "b", "c" should be "c", "b", and "a". 
Since "d" does not appear in S, it can be at any position in T. "dcba", "cdba", "cbda" are also valid outputs.

Note:
  • S has length at most 26, and no character is repeated in S.
  • T has length at most 200.
  • S and T consist of lowercase letters only.
Analysis:
1. count the occurrence of char in T.
2. loop through S, extract T's chars.
3. add remaining chars in T.
For instance, T: abcd, S: cba. While looping through S, we can know that cba are in T. Then add d anywhere.

Facebook has similar problem which uses number instead of character.

Solution:

class Solution {
    public String customSortString(String S, String T) {
        int[] count = new int[256];
        for (char c: T.toCharArray()) {
            count[c]++;
        }        

        StringBuilder sb =  new StringBuilder();
        for (char c: S.toCharArray()) {
            for (int i = 0; i < count[c]; i++) {
                sb.append(c);
            }
            count[c] = 0;
        }

        for (char c: T.toCharArray()) {
            if (count[c] != 0)
                sb.append(c);
        }

        return sb.toString();
    }
}

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