标签:char* parent code null sam pre 空间 str character
Given a string containing just the characters ‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘ and ‘]‘, determine if the input string is valid.An input string is valid if:
Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order.
Note that an empty string is also considered valid.
bool isValid(char* s) {
int k=0,i=0;
int p;
p=strlen(s);
int MAXSIZE=10000;//输入较大时,可能空间不足
char st;
st=(char)malloc(MAXSIZE*sizeof(char));
if(s==NULL)
return true;
if(p%2==1)
return false;
while(s[k]){
if((s[k]==‘{‘)||(s[k]==‘[‘)||(s[k]==‘(‘)){
st[i]=s[k];
++k;
++i;}
else{
switch(s[k]){
case ‘}‘:{
if(st[i-1]==‘{‘)//注意比较的数组以及序号
--i;
else
return false;
++k;
break;}
case ‘]‘:{
if(st[i-1]==‘[‘)
--i;
else
return false;
++k;
break;}
case ‘)‘:{
if(st[i-1]==‘(‘)
--i;
else
return false;
++k;
break;}
}
}
}
if(i==0)
return true;
else
return false;
}
标签:char* parent code null sam pre 空间 str character
原文地址:http://blog.51cto.com/13925487/2160956