算法练习Stack--P1021-easy

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1021. Remove Outermost Parentheses

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.

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.

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.

Return S after removing the outermost parentheses of every primitive string in the primitive decomposition of S.

Example 1:

Input: “(()())(())” Output: “()()()” Explanation: The input string is “(()())(())”, with primitive decomposition “(()())” + “(())”. After removing outer parentheses of each part, this is “()()” + “()” = “()()()”.

Example 2:

Input: “(()())(())(()(()))” Output: “()()()()(())” Explanation: The input string is “(()())(())(()(()))”, with primitive decomposition “(()())” + “(())” + “(()(()))”. After removing outer parentheses of each part, this is “()()” + “()” + “()(())” = “()()()()(())”.

Example 3:

Input: “()()” Output: "” Explanation: The input string is “()()”, with primitive decomposition “()” + “()”. After removing outer parentheses of each part, this is "” + "” = “”.

将最外层的括号删除,意味着只保留“最外层括号内的内容”。使用这个思路,需要判断什么时候出现了“最外层括号”。

顺序遍历,当左括号的个数与右括号的个数相同时,意味着可以“收集”括号中的内容了。然后开始下一个“括号内”内容。

所以需要记录三个变量:左括号个数、右括号个数、需要收集的内容的开始位置。

Runtime: 3 ms, faster than 69.58% of Java online submissions for Remove Outermost Parentheses. Memory Usage: 37 MB, less than 94.81% of Java online submissions for Remove Outermost Parentheses.

public String removeOuterParentheses(String S) {

    StringBuilder sb = new StringBuilder();
    int open=0, close=0, start=0;
    for(int i=0; i<S.length(); i++) {
        if(S.charAt(i) == '(') {
            open++;
        } else if(S.charAt(i) == ')') {
            close++;
        }
        if(open==close) {
            sb.append(S.substring(start+1, i));
            start=i+1;
        }
    }
    return sb.toString();
}
 
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