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 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 <= 100
s
contains only lower and upper case English letters.
给定一个字符串,如果相邻两个字符是同一字母的大小写形式,则需要删除这两个字母,问最终的字符串形式是什么。
简单题,使用栈,完整代码如下:
class Solution {
public:
string makeGood(string s) {
stack<char> stk;
for(int i = 0; i < s.size(); ++i) {
if(!stk.empty() && abs(stk.top() - s[i]) == 32) {
stk.pop();
} else {
stk.push(s[i]);
}
}
string ans="";
while(!stk.empty()) {
ans = string(1, stk.top()) + ans;
stk.pop();
}
return ans;
}
};
本代码提交AC。