leetcode:20. Valid Parentheses

描述

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.

思路

这里我有两个思路，思路一很快就实现了，并且效果还可以，思路二，在实现的时候出了一些问题。花费了不少时间，还是没有解决，为了刷题，先放下。

思路一：利用栈的思想。

思路二：想要利用hash表来减少判断的过程。不过出现了一下问题，没有跑起来。

代码

思路一：

class Solution {
public:
bool isValid(string s) {
int s_size = s.size();
vector<char> chsVec;
chsVec.push_back('a');
char temp;

bool chs_map[128] = {false};
chs_map['('] = true;
chs_map['{'] = true;
chs_map['['] = true;
chs_map[')'] = true;
chs_map['}'] = true;
chs_map[']'] = true;

for(int i = 0; i < s_size; i++){
temp = s[i];
if(chs_map[temp]){
switch(temp){
case '(':{
chsVec.push_back(temp);
break;
}
case '{':{
chsVec.push_back(temp);
break;
}
case '[':{
chsVec.push_back(temp);
break;
}
case ')':{
if(chsVec.back() == '('){
chsVec.pop_back();
}
else{
return false;
}
break;
}
case '}':{
if(chsVec.back() == '{'){
chsVec.pop_back();
}
else{
return false;
}
break;
}
case ']':{
if(chsVec.back() == '['){
chsVec.pop_back();
}
else{
return false;
}
break;
}

}
}
}
if(chsVec.size() == 1){
return true;
}
else{
return false;
}
}
};

思路二：

class Solution {
public:
bool isValid(string s) {
int s_size = s.size();
char chs_vec[s_size+1];
int index = 0;
char temp;

bool chs_map[128] = {false};
chs_map['('] = true;
chs_map['{'] = true;
chs_map['['] = true;
chs_map[')'] = true;
chs_map['}'] = true;
chs_map[']'] = true;

bool chs_left[128] = {false};
chs_map['('] = true;
chs_map['{'] = true;
chs_map['['] = true;

char chs_right[128];
chs_map[')'] = '(';
chs_map['}'] = '{';
chs_map[']'] = '[';

for(int i = 0; i < s_size; i++){
temp = s[i];
if(chs_map[temp]){
cout << temp << endl;
cout << chs_left[temp] << endl;
if(chs_left[temp]){
index++;
chs_vec[index] = temp;

}
else{
cout << index << endl;
cout << temp << endl;
if(chs_vec[index] == chs_right[temp]){
index--;
cout << index;
}
else{
return false;
}
}
}

}

cout << 'over' << endl;
if(index == 0){
return true;
}
else{
return false;
}
}
};

结果

他山之玉

C++ O(n) solutioin

int getV(char c)
{
if (c == '(') return 1;
else if (c == ')') return -1;
else if (c == '[') return 2;
else if (c == ']') return -2;
else if (c == '{') return 3;
else if (c == '}') return -3;
else return 0;
}

bool isValid(string s) {
stack<int> st;
for (int i = 0; i < s.size(); i++)
{
int v = getV(s[i]);
if (v < 0)
{
if (st.empty()) return false;
if (st.top() + v) return false;
st.pop();
}
else
{
st.push(v);
}
}

return st.empty();
}

这个思路将字符的比对转换成加减。

Java O(n) solution

public boolean isValid(String s) {
Stack<Character> stack = new Stack<Character>();
for (char c : s.toCharArray()) {
if (c == '(')
stack.push(')');
else if (c == '{')
stack.push('}');
else if (c == '[')
stack.push(']');
else if (stack.isEmpty() || stack.pop() != c)
return false;
}
return stack.isEmpty();
}

恩，这个更加简洁，值得借鉴。

python O(n) solution

class Solution:
# @return a boolean
def isValid(self, s):
stack = []
dict = {"]":"[", "}":"{", ")":"("}
for char in s:
if char in dict.values():
stack.append(char)
elif char in dict.keys():
if stack == [] or dict[char] != stack.pop():
return False
else:
return False
return stack == []