Leetcode Valid parentheses (C,c++,java,python)

Problem:

Given A string containing just the characters‘(‘,‘)‘,‘{‘,‘}‘,‘[‘and‘]‘, determine if the input string is valid.

the brackets must close in the correct order, " () "  and " () []{} "  are all valid But " (] "  and " ([)] "  are not.

Solution: Typical stack application, stack resolution, complexity O (n)
To give a string that contains only ‘(‘,  ‘)‘,  ‘{‘,  ‘}‘,  ‘[‘and the ‘]‘, the string that asks whether the decision is legal.
Java source code (spents 251ms):

public class Solution {public    Boolean isValid (String s) {        int length=s.length (), top=-1,index=0;        Char[] Stack=new char[length];        Char[] Str=s.tochararray ();        while (index<length) {            if (str[index]== ') ') {                if (top>=0 && stack[top]== ' (') top--;                else return false;            } else if (str[index]== '} ') {                if (top>=0 && stack[top]== ' {') top--;                else return false;            } else if (str[index]== '] ') {                if (top>=0 && stack[top]== ' [') top--;                else return false;            } else Stack[++top]=str[index];            index++;        }        return top==-1;}    }

C Language Source code (1MS):

BOOL IsValid (char* s) {    char stack[1000000];    int top=-1;    while (*s) {        if (*s== ') ') {            if (top>=0 && stack[top]== ' (') top--;            else return false;        } else if (*s== '} ') {            if (top>=0 && stack[top]== ' {') top--;            else return false;        } else if (*s== '] ') {            if (top>=0 && stack[top]== ' [') top--;            else return false;        } else stack[++top]=*s;        s++;    }    return top==-1;}

C + + source code (2ms):

Class Solution {public:    bool IsValid (string s) {        int top=-1,index=0,length=s.size ();        char* stack= (char*) malloc (sizeof (char) *length);        while (index<length) {            if (s[index]== ') ') {                if (top>=0 && stack[top]== ' (') top--;                else return false;            } else if (s[index]== '} ') {                if (top>=0 && stack[top]== ' {') top--;                else return false;            } else if (s[index]== '] ') {                if (top>=0 && stack[top]== ' [') top--;                else return false;            } else Stack[++top]=s[index];            index++;        }        return top==-1;}    ;

python source code (42MS):

Class solution:    # @param {string} s    # @return {boolean}    def isValid (self, s):        Length=len (s); top=-1; Index=0        stack=["For I in range (length)" While        index < length:            if s[index]== ') ':                if top>=0 and STA ck[top]== ' (': top-=1;                Else:return False            elif s[index]== '} ':                if top>=0 and stack[top]== ' {': top-=1;                Else:return False            elif s[index]== ':                if top>=0 and stack[top]== ' [': top-=1;                Else:return False            Else:top+=1;stack[top]=s[index]            index+=1        return top==-1

Leetcode Valid parentheses (C,c++,java,python)