LeetCode 1544. Make The String Great

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] and s[i + 1] where:

• 0 <= i <= s.length - 2
• s[i] is a lower-case letter and s[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.

Example 1:

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".

Example 2:

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" --> ""

Example 3:

Input: s = "s"
Output: "s"

Constraints:

• 1 <= s.length <= 100
• s contains only lower and upper case English letters.

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这题也是，如果不提前告诉我是stack的思想，我可能就不会做了，但是告诉我是stack以后就恍然大悟了，和1047挺像的，于是就借鉴了1047里用sb当作stack的方法写完了：

class Solution {
    public String makeGood(String s) {
        StringBuilder sb = new StringBuilder();
        for (char c : s.toCharArray()) {
            if (sb.length() != 0 && Character.toLowerCase(sb.charAt(sb.length() - 1)) == Character.toLowerCase(c) && sb.charAt(sb.length() - 1) != c) {
                sb.deleteCharAt(sb.length() - 1);
            } else {
                sb.append(c);
            }
        }
        return sb.toString();
    }
}

还有一点就是判断两个字符是不同大小写的，可以直接Math.abs(a - b) == 32

然后看了答案还有人用递归做了，妙啊，虽然还没有完全想通。

class Solution {
    public String makeGood(String s) {
        for (int i = 0; i < s.length() - 1; i++) {
            if (Math.abs(s.charAt(i) - s.charAt(i + 1)) == 32) {
                return makeGood(s.substring(0, i) + s.substring(i + 2, s.length()));
            }
        }
        return s;
    }
}