QUESTION:
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.
EXPLANATION:
easy的题目, 直接while循环, 如果最终结果没有变化, 那么说明已经不可以再优化.
SOLUTION:
class Solution {
public String makeGood(String s) {
int preLength = s.length();
int lastestLength = 0;
String result = s;
while (preLength != lastestLength) {
preLength = result.length();
for (int i = 0;i<result.length()-1;i++) {
if (Math.abs(result.charAt(i) - result.charAt(i+1)) == 32) {
String pre = "";
if (i != 0) {
pre = result.substring(0, i);
}
String last = "";
if (i+2 <= result.length() - 1) {
last = result.substring(i + 2, result.length());
}
result = pre+last;
break;
}
}
lastestLength = result.length();
}
return result;
}
}