the first one in Weekly Contest 201.
Difficulty : Easy
Related Topics : String、Stack
Given a string
s
of lower and upper case English letters.A good string is a string which doesn't have two adjacent characters
s[i]
ands[i + 1]
where:
0 <= i <= s.length - 2
s[i]
is a lower-case letter ands[i + 1]
is the same letter but in upper-case or vice-versa.To make the string good, you can choose two adjacent characters that make the string bad and remove them. You can keep doing this until the string becomes good.
Return the string after making it good. The answer is guaranteed to be unique under the given constraints.
Notice that an empty string is also good.
Input: s = "leEeetcode" Output: "leetcode" Explanation: In the first step, either you choose i = 1 or i = 2, both will result "leEeetcode" to be reduced to "leetcode".
Input: s = "abBAcC" Output: "" Explanation: We have many possible scenarios, and all lead to the same answer. For example: "abBAcC" --> "aAcC" --> "cC" --> "" "abBAcC" --> "abBA" --> "aA" --> ""
Input: s = "s" Output: "s"
1 <= s.length <= 100
s
contains only lower and upper case English letters.
- mine
- Java
Runtime: 2 ms, faster than 94.31%, Memory Usage: 39.6 MB, less than 62.22% of Java online submissions
// O(N)time // O(N)space public String makeGood(String s) { char[] arr = s.toCharArray(); int n = arr.length; LinkedList<Character> list = new LinkedList<>(); for(int i = 0; i < n; i++){ if(!list.isEmpty()){ if(Math.abs(list.getLast() - arr[i]) == 'a' - 'A'){ list.removeLast(); continue; } } if(i + 1 != n && Math.abs(arr[i + 1] - arr[i]) == 'a' - 'A') i++; else list.add(arr[i]); } StringBuilder sb = new StringBuilder(); while(!list.isEmpty()){ sb.append(list.removeFirst()); } return sb.toString(); }
- Java