1544. Make The String Great
- Hardness: \(\color{green}\textsf{Easy}\)
- Ralated Topics:
String、Stack
一、題目
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 - 2s[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.
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 <= 100scontains only lower and upper case English letters.
二、分析
- 當前後兩個字元互為大小寫的字元時，將之移除。可利用
stack來操作之。
三、解題
1. Stack
- Time complexity: \(O(n)\)
- Space complexity: \(O(1)\)
char trans(char c) { // 大小寫互換
return ('a' <= c && c <= 'z') ? c - 32 : c + 32;
}
string makeGood(string s) {
string st;
for (char c : s) {
if (st.empty()) {
st.push_back(c);
} else {
if (st.back() == trans(c)) { // 當 stack 頂端與下一個字元互為大小寫
st.pop_back();
} else {
st.push_back(c);
}
}
}
return st;
}