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🐱(string): 771. 宝石与石头

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`@@ -1423,6 +1423,49 @@ class Solution(object):`
`1423`	`1423`	`return result`
`1424`	`1424`	```
`1425`	`1425`
	`1426`	`+## 771. 宝石与石头`
	`1427`	`+`
	`1428`	`+[原题链接](https://leetcode-cn.com/problems/jewels-and-stones/)`
	`1429`	`+`
	`1430`	`+### 解法一:遍历字符串`
	`1431`	`+`
	`1432`	`+遍历字符串,寻找 S 中存在于 J 中的字母个数。`
	`1433`	`+`
	`1434`	+```python
	`1435`	`+class Solution:`
	`1436`	`+ def numJewelsInStones(self, J: str, S: str) -> int:`
	`1437`	`+ res = 0`
	`1438`	`+ for j in J:`
	`1439`	`+ for s in S:`
	`1440`	`+ if j == s:`
	`1441`	`+ res += 1`
	`1442`	`+ return res`
	`1443`	+```
	`1444`	`+`
	`1445`	`+pythonic 一点:`
	`1446`	`+`
	`1447`	+```python
	`1448`	`+class Solution:`
	`1449`	`+ def numJewelsInStones(self, J: str, S: str) -> int:`
	`1450`	`+ return len([x for x in S if x in J])`
	`1451`	+```
	`1452`	`+`
	`1453`	`+### 解法二:哈希表`
	`1454`	`+`
	`1455`	+记录 `S` 中每个字母出现的次数,再把 `J` 中存在的字母次数都相加。
	`1456`	`+`
	`1457`	+```python
	`1458`	`+class Solution:`
	`1459`	`+ def numJewelsInStones(self, J: str, S: str) -> int:`
	`1460`	`+ s_count = dict()`
	`1461`	`+ for s in S:`
	`1462`	`+ s_count[s] = s_count.get(s, 0) + 1`
	`1463`	`+ res = 0`
	`1464`	`+ for j in J:`
	`1465`	`+ res += s_count.get(j, 0)`
	`1466`	`+ return res`
	`1467`	+```
	`1468`	`+`
`1426`	`1469`	`## 1108. IP 地址无效化`
`1427`	`1470`
`1428`	`1471`	`[原题链接](https://leetcode-cn.com/problems/defanging-an-ip-address/)`

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