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🐱(string): 848. 字母移位

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`@@ -1466,6 +1466,39 @@ class Solution:`
`1466`	`1466`	`return res`
`1467`	`1467`	```
`1468`	`1468`
	`1469`	`+## 848. 字母移位`
	`1470`	`+`
	`1471`	`+[原题链接](https://leetcode-cn.com/problems/shifting-letters/)`
	`1472`	`+`
	`1473`	`+### 思路`
	`1474`	`+`
	`1475`	`+越靠前的字母需要移动次数越多,可以算出每个字母需要移动的总次数,然后再对字母进行统一移动处理。时间复杂度为 $O(n)$。`
	`1476`	`+`
	`1477`	+```python
	`1478`	`+class Solution:`
	`1479`	`+ def shiftingLetters(self, S: str, shifts: List[int]) -> str:`
	`1480`	`+ length = len(shifts)`
	`1481`	`+`
	`1482`	`+ # 每个字母要移动的总次数`
	`1483`	`+ num = [0 for _ in range(length)]`
	`1484`	`+ num[length - 1] = shifts[length - 1]`
	`1485`	`+ for i in range(length - 2, -1, -1):`
	`1486`	`+ num[i] = shifts[i] + num[i + 1]`
	`1487`	`+`
	`1488`	`+ s_list = list(S)`
	`1489`	`+ ord_a = ord('a')`
	`1490`	`+ ord_z = ord('z')`
	`1491`	`+ for i in range(length):`
	`1492`	`+ shift = num[i] % 26`
	`1493`	`+ c = s_list[i]`
	`1494`	`+ ord_res = ord(c) + shift`
	`1495`	`+ if ord_res > ord_z:`
	`1496`	`+ ord_res = ord_res - ord_z - 1 + ord_a`
	`1497`	`+ s_list[i] = chr(ord_res)`
	`1498`	`+`
	`1499`	`+ return "".join(s_list)`
	`1500`	+```
	`1501`	`+`
`1469`	`1502`	`## 1108. IP 地址无效化`
`1470`	`1503`
`1471`	`1504`	`[原题链接](https://leetcode-cn.com/problems/defanging-an-ip-address/)`

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