LeetCode 32 Longest Valid Parentheses (C,C++,Java,Python)

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Problem:

Given a string containing just the characters ‘(‘ and ‘)‘, find the length of the longest valid (well-formed) parentheses substring.

For "(()", the longest valid parentheses substring is "()", which has length = 2.

Another example is ")()())", where the longest valid parentheses substring is "()()", which has length = 4.

Solution:維持一個字元棧和位置棧，字元棧記錄沒有匹配的字元，位置棧記錄每一個入棧的字元的位置，如果入棧的是右括弧，並且棧頂為左括弧，則退棧，求出當前連續退棧的字元長度，求出這個最大值就是。
題目大意：給一個字串，只包含左括弧和右括弧，求出這個字串合法括弧的最大長度。
Java原始碼（350ms）：

public class Solution {    public int longestValidParentheses(String s) {        int len=s.length(),max=0;        Stack<Character> str_stack = new Stack<Character>();        Stack<Integer> pos_stack = new Stack<Integer>();        char[] str=s.toCharArray();        for(int i=0;i<len;i++){            if(str[i]=='('){                str_stack.push('(');                pos_stack.push(i);            }else{                if(str_stack.size()>0 && str_stack.peek().equals('(')){                    str_stack.pop();                    pos_stack.pop();                    int tmp=pos_stack.size()==0?i+1:i-pos_stack.peek();                    max=Math.max(max,tmp);                }else{                    str_stack.push(')');                    pos_stack.push(i);                }            }        }        return max;    }}

C語言原始碼（3ms）：

int longestValidParentheses(char* s) {    int len=strlen(s),max=0,top=0,i,tmp;    int* pos_stack=(int*)malloc(sizeof(int)*len);    char* str_stack=(char*)malloc(sizeof(char)*len);    for(i=0;s[i];i++){        if(s[i]=='('){            str_stack[top]='(';            pos_stack[top]=i;            top++;        }else{            if(str_stack[top-1]=='('){                top--;                if(top==0)tmp=i+1;                else tmp=i-pos_stack[top-1];                max=max>tmp?max:tmp;            }else{                str_stack[top]=')';                pos_stack[top]=i;                top++;            }        }    }    return max;}

C++原始碼（7ms）：

class Solution {public:    int longestValidParentheses(string s) {        int max=0,top=0,len=s.size();        int* pos_stack=(int*)malloc(sizeof(int)*len);        char* str_stack=(char*)malloc(sizeof(char)*len);        for(int i=0;i<len;i++){            if(s[i]=='('){                str_stack[top]='(';                pos_stack[top]=i;                top++;            }else{                if(str_stack[top-1]=='('){                    int tmp;                    top--;                    if(top==0)tmp=i+1;                    else tmp=i-pos_stack[top-1];                    max=max>tmp?max:tmp;                }else{                    str_stack[top]=')';                    pos_stack[top]=i;                    top++;                }            }        }        return max;    }};

Python原始碼（105ms）：

class Solution:    # @param {string} s    # @return {integer}    def longestValidParentheses(self, s):        length=len(s);top=0;Max=0        str_stack=[' ' for i in range(length)]        pos_stack=[0 for i in range(length)]        for i in range(length):            if s[i]=='(':                str_stack[top]='('                pos_stack[top]=i                top+=1            else:                if top>0 and str_stack[top-1]=='(':                    top-=1                    tmp=i+1 if top==0 else i-pos_stack[top-1]                    Max=Max if Max>tmp else tmp                else:                    str_stack[top]=')'                    pos_stack[top]=i                    top+=1        return Max

LeetCode 32 Longest Valid Parentheses (C,C++,Java,Python)