描述
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"
Note:
1 <= s.length <= 100
s contains only lower and upper case English letters.
解析
根据题意，只需要将字符串 s 中的相同字母但是大小写不同的连着的两个字母都剔除返回剩下的字符串即可，思路简单，就是遍历 s 中的字符，然后当列表 r 不为空且 r 中最后一个元素于字符 c 的 ascii 距离为 32 ，则执行 r.pop(-1) ，否则将 c 追加到 r 中，最后将 r 转化为 字符串即可。
解答
class Solution(object):
def makeGood(self, s):
"""
:type s: str
:rtype: str
"""
r = []
for c in s:
if r and abs(ord(r[-1])-ord(c))==32:
r.pop(-1)
else:
r.append(c)
return ''.join(r)
运行结果
Runtime: 24 ms, faster than 76.34% of Python online submissions for Make The String Great.
Memory Usage: 13.6 MB, less than 44.27% of Python online submissions for Make The String Great.
原题链接：leetcode.com/problems/ma…
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