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Merge pull request youngyangyang04#2799 from gazeldx/E0232

0232.用栈实现队列.md 时间复杂度应为O(1)

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`@@ -113,7 +113,7 @@ public:`
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`114`	`114`	```
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`116`		`-* 时间复杂度: push和empty为O(1), pop和peek为O(n)`
	`116`	`+* 时间复杂度: 都为O(1)。pop和peek看起来像O(n),实际上一个循环n会被使用n次,最后还是O(1)。`
`117`	`117`	`* 空间复杂度: O(n)`
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`@@ -77,7 +77,7 @@ for (int i = 0; i < array.size(); i++) {`
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`78`	`78`	`时间复杂度可以做到O(n^2),但还是比较费时的,因为不好做剪枝操作。`
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`80`		`-所以这道题目使用双指针法才是最为合适的,用双指针做这道题目才能就能真正体会到,通过前后两个指针不算向中间逼近,在一个for循环下完成两个for循环的工作。`
	`80`	`+所以这道题目使用双指针法才是最为合适的,用双指针做这道题目才能就能真正体会到,通过前后两个指针不断向中间逼近,在一个for循环下完成两个for循环的工作。`
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`82`	`82`	`只用双指针法时间复杂度为O(n^2),但比哈希法的O(n^2)效率高得多,哈希法在使用两层for循环的时候,能做的剪枝操作很有限。`
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