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🐱(double-pointer): 392. 判断子序列

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`27`	`27`	`* [位运算](algorithm/bit/)`
`28`	`28`	`* [数学题](algorithm/math/)`
`29`	`29`	`* [回溯](algorithm/backtrack/)`
	`30`	`+ * [双指针](algorithm/double-pointer/)`
`30`	`31`	`* [其他](algorithm/other/)`
`31`	`32`	`* 周赛`
`32`	`33`	`* [第 121 场周赛](weekly/121/)`

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	`1`	`+## 392. 判断子序列`
	`2`	`+`
	`3`	`+[原题链接](https://leetcode-cn.com/problems/is-subsequence/)`
	`4`	`+`
	`5`	`+### 思路`
	`6`	`+`
	`7`	`+双指针。`
	`8`	`+`
	`9`	+指针 `i` 指向 `s` 第一个字符,指针 `j` 指向 `t` 第一个字符。逐一判断 `i` 所指向的字符是否在 `t` 中存在。
	`10`	`+`
	`11`	+- 如果 `s[i] != t[j]`:`j = j + 1`,继续比对下一个字符
	`12`	+- 如果 `s[i] == t[j]`:`i = i + 1`,`j = j + 1`,继续进行下一个 `s[i]` 的查找
	`13`	`+`
	`14`	+```python
	`15`	`+class Solution(object):`
	`16`	`+ def isSubsequence(self, s, t):`
	`17`	`+ """`
	`18`	`+ :type s: str`
	`19`	`+ :type t: str`
	`20`	`+ :rtype: bool`
	`21`	`+ """`
	`22`	`+ s_length = len(s)`
	`23`	`+ t_length = len(t)`
	`24`	`+`
	`25`	`+ i = 0`
	`26`	`+ j = 0`
	`27`	`+`
	`28`	`+ while i < s_length:`
	`29`	`+ s_c = s[i]`
	`30`	`+ has = False`
	`31`	`+ while j < t_length:`
	`32`	`+ t_c = t[j]`
	`33`	`+ j += 1`
	`34`	`+ if s_c == t_c:`
	`35`	`+ i += 1`
	`36`	`+ has = True`
	`37`	`+ break`
	`38`	`+`
	`39`	`+ if has == False:`
	`40`	`+ return False`
	`41`	`+`
	`42`	`+ return True`
	`43`	+```

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