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Commit ded538a

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1021. Remove Outermost Parentheses
Signed-off-by: Zhaowei Fang <zwfangio@gmail.com>
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‎README.md‎

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* [345. Reverse Vowels of a String](src/0345_reverse_vowels_of_a_string/reverse_vowels.go)&nbsp;&nbsp;&nbsp;*`string;`*&nbsp;&nbsp;*`double index`*
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* [438. Find All Anagrams in a String](src/0438_all_anagrams_in_a_string/all_anagrams_in_a_string.go)&nbsp;&nbsp;&nbsp;*`sliding window`*
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* [557. Reverse Words in a String III](src/0557_reverse_words_in_a_string_3/reverse_words_in_a_string_3.go)
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* [1021. Remove Outermost Parentheses](src/1021_Remove_Outermost_Parentheses/remove_outmost_parentheses.go)&nbsp;&nbsp;&nbsp;*`stack`*
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### Linked List
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* [2. Add Two Numbers](src/0002_add_two_numbers/add_two_numbers.go)&nbsp;&nbsp;&nbsp;*`recursion;`*&nbsp;&nbsp;*`math`*
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/*
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1021. Remove Outermost Parentheses
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A valid parentheses string is either empty (""), "(" + A + ")", or A + B, where A and B are valid parentheses strings, and + represents string concatenation. For example, "", "()", "(())()", and "(()(()))" are all valid parentheses strings.
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A valid parentheses string S is primitive if it is nonempty, and there does not exist a way to split it into S = A+B, with A and B nonempty valid parentheses strings.
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Given a valid parentheses string S, consider its primitive decomposition: S = P_1 + P_2 + ... + P_k, where P_i are primitive valid parentheses strings.
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Return S after removing the outermost parentheses of every primitive string in the primitive decomposition of S.
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Example 1:
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Input: "(()())(())"
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Output: "()()()"
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Explanation:
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The input string is "(()())(())", with primitive decomposition "(()())" + "(())".
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After removing outer parentheses of each part, this is "()()" + "()" = "()()()".
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Example 2:
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Input: "(()())(())(()(()))"
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Output: "()()()()(())"
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Explanation:
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The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))".
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After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())".
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Example 3:
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Input: "()()"
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Output: ""
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Explanation:
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The input string is "()()", with primitive decomposition "()" + "()".
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After removing outer parentheses of each part, this is "" + "" = "".
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Note:
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S.length <= 10000
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S[i] is "(" or ")"
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S is a valid parentheses string
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*/
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// time: 2020年05月04日
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package removeoutmostparentheses
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// time complexity: O(n), where n is length of S.
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// space complexity: O(1)
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func removeOuterParentheses(S string) string {
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// S is valid parentheses string, so, s is empty "" or starts with "("
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if len(S) == 0 {
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return S
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}
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var stackLength, left int
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var ret string
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for i := 0; i < len(S); i++ {
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if stackLength == 0 {
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left = i
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}
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if S[i] == '(' {
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stackLength++
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} else {
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stackLength--
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}
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if stackLength == 0 {
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ret += S[left+1 : i]
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}
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}
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return ret
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}
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package removeoutmostparentheses
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import (
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"testing"
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)
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func TestRemoveOuterParentheses(t *testing.T) {
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testCases := []map[string]string{
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{"case": "(()())(())", "expected": "()()()"},
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{"case": "(()())(())(()(()))", "expected": "()()()()(())"},
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{"case": "", "expected": ""},
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{"case": "()()", "expected": ""},
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}
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for _, testCase := range testCases {
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if testCase["expected"] != removeOuterParentheses(testCase["case"]) {
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t.Errorf("hello")
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}
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}
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}

‎src/README.md‎

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|0728|[Self Dividing Numbers](./0728_self_dividing_numbers/self_dividing_numbers.go)|Easy||
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|0735|[735. Asteroid Collision](0735_asteroid_collision/ac.go)|Medium|*`stack`*|
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|0747|[Largest Number At Least Twice of Others](./0747_largest_number_at_least_twice_of_others/largest_number_at_least_twice_of_others.go)|Easy||
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|0872|[872. Leaf-Similar Trees](0872_leaf_similar_trees/leaf_similar_trees.go)|Easy|*`binary tree`*|
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|0872|[872. Leaf-Similar Trees](0872_leaf_similar_trees/leaf_similar_trees.go)|Easy|*`binary tree`*|
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|1021|[1021. Remove Outermost Parentheses](1021_Remove_Outermost_Parentheses/remove_outmost_parentheses.go)|Easy|*`stack`*|

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