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更新+补充 1047.删除字符串中的所有相邻重复项.md python代码

补充使用双指针的方法,但是更推荐栈的做法

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Original file line number	Diff line number	Diff line change
`@@ -197,15 +197,38 @@ class Solution {`
`197`	`197`
`198`	`198`	`Python:`
`199`	`199`	```python3
	`200`	`+# 方法一,使用栈,推荐!`
`200`	`201`	`class Solution:`
`201`	`202`	`def removeDuplicates(self, s: str) -> str:`
`202`		`- t = list()`
`203`		`- for i in s:`
`204`		`- if t and t[-1] == i:`
`205`		`- t.pop(-1)`
	`203`	`+ res = list()`
	`204`	`+ for item in s:`
	`205`	`+ if res and res[-1] == item:`
	`206`	`+ t.pop()`
`206`	`207`	`else:`
`207`		`- t.append(i)`
`208`		`- return "".join(t) # 字符串拼接`
	`208`	`+ t.append(item)`
	`209`	`+ return "".join(res) # 字符串拼接`
	`210`	+```
	`211`	`+`
	`212`	+```python3
	`213`	`+# 双指针`
	`214`	`+class Solution:`
	`215`	`+ def removeDuplicates(self, s: str) -> str:`
	`216`	`+ res = list(s)`
	`217`	`+ slow = fast = 0`
	`218`	`+ length = len(res)`
	`219`	`+`
	`220`	`+ while fast < length:`
	`221`	`+ # 如果一样直接换,不一样会把后面的填在slow的位置`
	`222`	`+ res[slow] = res[fast]`
	`223`	`+`
	`224`	`+ # 如果发现和前一个一样,就退一格指针`
	`225`	`+ if slow > 0 and res[slow] == res[slow - 1]:`
	`226`	`+ slow -= 1`
	`227`	`+ else:`
	`228`	`+ slow += 1`
	`229`	`+ fast += 1`
	`230`	`+`
	`231`	`+ return ''.join(res[0: slow])`
`209`	`232`	```
`210`	`233`
`211`	`234`	`Go:`

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