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L20.java
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74 lines (66 loc) · 2.16 KB
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package com.liang.leetcode.stack;
import java.util.HashMap;
import java.util.Map;
import java.util.Stack;
/**
* @ClassName L20
* @description 有效的括号
* @Author LiaNg
* @Date 2019-11-20
*/
public class L20 {
public static void main(String[] args) {
String s = "[";
L20 l20 = new L20();
System.out.println("l20.isValid(s) = " + l20.isValid(s));
}
/**
* 给定一个只包括 '(',')','{','}','[',']' 的字符串,判断字符串是否有效。
* 有效字符串需满足:
* 左括号必须用相同类型的右括号闭合。
* 左括号必须以正确的顺序闭合。
* 注意空字符串可被认为是有效字符串。
* 来源:力扣(LeetCode)
* 链接:https://leetcode-cn.com/problems/valid-parentheses
* 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
*/
public boolean isValid(String s) {
Map<Character, Character> map = new HashMap<>(3);
map.put(')', '(');
map.put('}', '{');
map.put(']', '[');
Stack<Character> stack = new Stack<>();
for (char c : s.toCharArray()) {
if (!map.containsKey(c)) {
stack.push(c);
} else if (!stack.empty() && map.get(c).equals(stack.peek())) {
stack.pop();
} else {
return false;
}
}
return stack.empty();
}
/**
* LeetCode 耗时最短解答
*/
public boolean isValid1(String s) {
if ("".equals(s)) {
return true;
}
char[] stack = new char[s.length()];
int stackCur = -1;
char[] split = s.toCharArray();
for (char cur : split) {
if (cur == '{' || cur == '(' || cur == '[') {
stack[++stackCur] = cur;
} else if (stackCur == -1
|| cur == ')' && stack[stackCur--] != '('
|| cur == ']' && stack[stackCur--] != '['
|| cur == '}' && stack[stackCur--] != '{') {
return false;
}
}
return stackCur == -1;
}
}