给出一个仅包含字符'(',')','{','}','['和']',的字符串，判断给出的字符串是否是合法的括号序列
括号必须以正确的顺序关闭，"()"和"()[]{}"都是合法的括号序列，但"(]"和"([)]"不合法。
括号必须以正确的顺序关闭，"()"和"()[]{}"都是合法的括号序列，但"(]"和"([)]"不合法。
数据范围：字符串长度 
要求：空间复杂度 )
，时间复杂度
[编程题]有效括号序列
import java.util.*;
public class Solution {
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
*
* @param s string字符串
* @return bool布尔型
*/
public boolean isValid (String s) {
// write code here
String[] split = s.split(",");
int length = split.length;
if (length == 1 ) {
return false;
} else {
if (length % 2 != 0) {
return false;
} else {
boolean flag = true;
int i = 0;
while (flag) {
if (split[i] == "(") {
if (split[i + 1] == ")") {
flag = true;
} else flag = false;
} else if (split[i] == "[") {
if (split[i + 1] == "]") {
flag = true;
} else flag = false;
} else if (split[i] == "{") {
if (split[i + 1] == "}") {
flag = true;
} else flag = false;
} else flag = false;
}
return flag;
}
}
}
}
编辑于 2024-03-26 16:20:34
回复(0)
import java.util.*;
public class Solution {
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
*
* @param s string字符串
* @return bool布尔型
*/
public boolean isValid (String s) {
Deque<Character> deque = new ArrayDeque<Character>();
// s = s.substring(1, s.length() - 1);
System.out.println(s);
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (c == '(' || c == '[' || c == '{') {
deque.push(c);
} else {
if (deque.size() == 0) {
return false;
}
char c1 = deque.pop();
String str = c1 + "" + c;
if (!(str.equals("()") || str.equals("[]") || str.equals("{}"))) {
return false;
}
}
}
return deque.size() == 0 ? true : false;
}
}
编辑于 2023-12-08 00:25:20
回复(1)
import java.util.*;
public class Solution {
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
*
* @param s string字符串
* @return bool布尔型
*/
public boolean isValid (String s) {
// write code here
Map<Character, Character> map = new HashMap<>();
map.put(')', '(');
map.put(']', '[');
map.put('}', '{');
Stack<Character> signs = new Stack<>();
for (int i = 0; i < s.length(); i++) {
if (signs.empty() || map.values().contains(s.charAt(i))) { // open sign
signs.push(s.charAt(i));
} else { // close sign
if (signs.peek() == map.get(s.charAt(i))) {
signs.pop();
}
}
}
return signs.empty() ? true : false;
}
}
发表于 2023-11-19 17:54:18
回复(1)
public boolean isValid (String s) {
Stack<Character> stack = new Stack<>();
for (int i = 0; i < s.length(); i++) {
char ch = s.charAt(i);
if (stack.isEmpty()) {
stack.push(ch);
} else {
char top = stack.peek();
if ((top == '(' && ch == ')') || (top == '{' && ch == '}') || (top == '[' &&
ch == ']')) {
stack.pop();
} else {
stack.push(ch);
}
}
}
return stack.isEmpty();
}
发表于 2023-09-23 19:59:07
回复(0)
本来是准备暴力解决的，但是突然看到有人使用栈，想到这个题明显和栈很搭，所以重新用栈写了一遍回答
import java.util.*;
public class Solution {
class AAAA {
char c;
AAAA next;
AAAA pri;
public AAAA (char c) {
this.c = c;
}
}
AAAA quene;
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
*
* @param s string字符串
* @return bool布尔型
*/
public boolean isValid (String s) {
// write code here
int lenght = s.length();
quene = new AAAA('0');
for (int i = 0; i < lenght ; i++) {
char c = s.charAt(i);
if (c == '(' || c == '[' || c == '{') {
// 入栈
addAAAA(c);
// 出栈并检查
} else if (quene.c == '0' || !check(c)) {
return false;
}
}
return quene.c == '0';
}
public boolean check (char c) {
// ')' - '(' == 1;
// ']' - '[' == 2;
// '}' - '{' == 2
char a = outAAAA();
return (c - a < 3) && (c - a > 0);
}
public boolean addAAAA (char c) {
AAAA q = new AAAA(c);
q.pri = quene;
quene.next = q;
quene = q;
return true;
}
// 出栈
public char outAAAA() {
char c = quene.c;
AAAA q = quene.pri;
quene.pri = null;
quene = q;
quene.next = null;
return c;
}
}
发表于 2023-08-19 17:14:56
回复(0)
import java.util.*;
public class Solution {
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
*
* @param s string字符串
* @return bool布尔型
*/
public boolean isValid (String s) {
// write code here
Stack<Character> s1 = new Stack<Character>();
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == '{' || s.charAt(i) == '[' || s.charAt(i) == '(') {
s1.push(s.charAt(i));
} else {
if (!s1.isEmpty() && isMatch(s1.peek(), s.charAt(i))) {
s1.pop();
} else {
return false;
}
}
}
if (s1.isEmpty()) return true;
return false;
}
public boolean isMatch (char s1, char s2) {
switch (s1) {
case '{':
return '}' == s2 ? true : false;
case '[':
return ']' == s2 ? true : false;
case '(':
return ')' == s2 ? true : false;
}
return false;
}
}
发表于 2023-08-12 16:27:07
回复(0)
import java.util.*;
public class Solution {
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
*
* @param s string字符串
* @return bool布尔型
*/
public boolean isValid (String s) {
// write code here
if (s.length() == 0 || s.length() % 2 == 1) {
return false;
}
if (s.charAt(0) == ')' || s.charAt(0) == '}' || s.charAt(0) == ']') {
return false;
}
Stack<Character> stack = new Stack<>();
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == '(' || s.charAt(i) == '{' || s.charAt(i) == '[') {
stack.push(s.charAt(i));
} else {
Character pop = stack.pop();
if (s.charAt(i) == ')' && pop != '(') {
return false;
}
if (s.charAt(i) == '}' && pop != '{') {
return false;
}
if (s.charAt(i) == ']' && pop != '[') {
return false;
}
}
}
return stack.size() == 0;
}
}
发表于 2023-08-11 23:01:28
回复(0)
public class Solution {
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
*
* @param s string字符串
* @return bool布尔型
*/
public static boolean isValid (String s) {
// write code here
if ((s.length() % 2) != 0) {
return false;
}
int i = 0;
StringBuilder sb = new StringBuilder(s);
while (s.length() >= 2) {
if ((s.charAt(i) == '(') && (s.charAt(i + 1) == ')')) {
sb.delete(i, i + 2);
s = sb.toString();
i = 0;
continue;
}
if ((s.charAt(i) == '[') && (s.charAt(i + 1) == ']')) {
sb.delete(i, i + 2);
s = sb.toString();
i = 0;
continue;
}
if ((s.charAt(i) == '{') && (s.charAt(i + 1) == '}')) {
sb.delete(i, i + 2);
s = sb.toString();
i = 0;
continue;
}
i++;
if (i == s.length() - 1) {
return false;
}
}
return true;
}
}
发表于 2023-08-05 17:29:47
回复(0)
public boolean isValid (String s) {
Stack<Character> st=new Stack<>();
for(int i=0;i<s.length();i++){
if(s.charAt(i)=='('){
st.push(')');
}else if(s.charAt(i)=='['){
st.push(']');
}else if(s.charAt(i)=='{'){
st.push('}');
}else{ //不是增加元素，是要弹出，但st为空，那就false
if(st.isEmpty()){
return false;
}
if(st.pop()!=s.charAt(i)){
return false;
}
}
}
if(!st.isEmpty()){
return false;
}
return true;
}
发表于 2023-07-19 10:58:36
回复(0)
import java.util.*;
public class Solution {
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
*
* @param s string字符串
* @return bool布尔型
*/
public boolean isValid (String s) {
// write code here
if(s == null || s.length() == 0) {
return true;
}
int len = s.length();
if(len % 2 != 0) {
return false;
}
char[] chs = s.toCharArray();
Stack<Character> sta = new Stack<>();
for(int i = 0; i < chs.length; i++) {
if(chs[i] == '(' || chs[i] == '{' || chs[i] == '[') {
sta.push(chs[i]);
} else if(sta.isEmpty()) {
return false;
} else {
char ch = sta.peek();
if(ch == '(' && chs[i] == ')') {
sta.pop();
} else if(ch == '{' && chs[i] == '}') {
sta.pop();
} else if(ch == '[' && chs[i] == ']') {
sta.pop();
} else {
return false;
}
}
}
return sta.isEmpty();
}
}
发表于 2023-07-15 16:13:38
回复(0)
import java.util.*;
public class Solution {
/**
*
* @param s string字符串
* @return bool布尔型
*/
public boolean isValid (String s) {
// write code here
String[] str=s.split("");
int flag=0;
for(int i=0;i<str.length/2;i++){
if((str[i].equals("(")&&str[str.length-i-1].equals(")")) || (str[i].equals("[")&&str[str.length-i-1].equals("]")
)|| (str[i].equals("{")&&str[str.length-i-1].equals("}"))){
flag=1;
}
else {
flag=0;
}
}
if(flag==1) return true;
return false;
}
}
发表于 2023-06-14 17:34:08
回复(0)
使用stack，解决问题很容易
import java.util.*;
public class Solution {
/**
*
* @param s string字符串
* @return bool布尔型
*/
public boolean isValid(String s) {
Stack<Character> stack = new Stack<>();
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (!stack.isEmpty()) {
if (c == ')' && stack.pop() == '(') {
continue;
}
if (c == ']' && stack.pop() == '[') {
continue;
}
if (c == '}' && stack.pop() == '{') {
continue;
}
}
stack.add(c);
}
return stack.isEmpty();
}
}
发表于 2023-05-16 22:47:21
回复(2)
import java.util.*;
//出栈入栈 public class Solution {
/**
*
* @param s string字符串
* @return bool布尔型
*/
public boolean isValid (String s) {
char[] c = s.toCharArray();
Stack<Character> stack = new Stack<>();
if (c.length % 2 != 0) return false;
for (int i = 0; i < c.length; i++) {
if (c[i] == '[') stack.push(']');
else if (c[i] == '{') stack.push('}');
else if (c[i] == '(') stack.push(')');
else if (stack.isEmpty() || c[i] != stack.peek()) return false;
else stack.pop();
}
return stack.isEmpty();
}
}
发表于 2023-03-26 16:24:26
回复(0)
public boolean isValid (String s) {
// write code here
//利用哈希存储字符，避免if语句中出现冗余比较语句
HashMap<Character, Character> map = new HashMap<Character, Character>() {
{
put(')', '(');
put(']', '[');
put('}', '{');
}
};
Stack<Character> stack = new Stack<>();
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (map.containsKey(c)) {
if (stack.isEmpty()) return false;
if (map.get(c) != stack.peek()) return false;
stack.pop();
} else {
stack.push(c);
}
}
return stack.isEmpty();
}
发表于 2023-03-20 20:13:26
回复(1)
为什么这个会爆数组下标异常啊？？？求大佬详解
public class Solution {
/**
*
* @param s string字符串
* @return bool布尔型
*/
public boolean isValid (String s) {
Stack<Character> sin = new Stack<Character>();
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (c == '[' || c == '(' || c == '{') {
sin.push(c);
} else if (c == ']' && sin.peek() == '[') {
sin.pop();
} else if (c == ')' && sin.peek() == '(') {
sin.pop();
} else if (c == '}' && sin.peek() == '{') {
sin.pop();
} else return false;
}
return sin.isEmpty();
}
}
发表于 2023-03-20 16:28:50
回复(1)
import java.util.*;
public class Solution {
/**
*
* @param s string字符串
* @return bool布尔型
*/
public boolean isValid (String s) {
Stack<Character> stack = new Stack<>();
int i = 0, len = s.length();
for (i = 0; i < len; i++) {
switch (s.charAt(i)) {
case '{' :
case '[' :
case '(' :
stack.push(s.charAt(i));
break;
case '}' :
if (stack.isEmpty() || stack.peek() != '{') {
return false;
}
stack.pop();
break;
case ']' :
if (stack.isEmpty() || stack.peek() != '[') {
return false;
}
stack.pop();
break;
case ')' :
if (stack.isEmpty() || stack.peek() != '(') {
return false;
}
stack.pop();
break;
}
}
return stack.isEmpty();
}
}
发表于 2023-02-04 18:45:16
回复(0)
public class Solution {
public boolean isValid (String s) {
if(s==null||s.length()==0){
return false;
}
Stack<Character> stack = new Stack<>();
for(int i=0;i<s.length();i++){
if(s.charAt(i)==')'){
if(stack.isEmpty()||stack.pop()!='('){
return false;
}
}else if(s.charAt(i)=='}'){
if(stack.isEmpty()||stack.pop()!='{'){
return false;
}
}else if(s.charAt(i)==']'){
if(stack.isEmpty()||stack.pop()!='['){
return false;
}
}else{
stack.push(s.charAt(i));
}
}
return stack.isEmpty();
}
}
发表于 2023-02-01 17:00:34
回复(0)
public boolean isValid (String s) {
Stack<Character> s1 = new Stack<>();
String mod_left = "[{(";
String mod_right = "]})";
for (int i = 0; i < s.length(); i++) {
if (mod_left.contains(String.valueOf(s.charAt(i)))) {
s1.add(s.charAt(i));
continue;
}
if (mod_right.contains(String.valueOf(s.charAt(i))) && !s1.empty()) {
Character pop = s1.pop();
if (s.charAt(i) == ']' && pop == '[' || s.charAt(i) == '}' && pop == '{' ||
s.charAt(i) == ')' && pop == '(')
continue;
else
return false;
} else
return false;
}
return s1.empty();
}
发表于 2023-01-01 15:12:43
回复(0)
问题信息
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