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题目
判断字符串中的括号是否有效。要求括号成对出现,并且括号顺序对应上。例如:[12(fgsf)4]-有效、{d[]df34}-有效、{f3[aer)}-无效、{3)32}-无效。
分析
依然考察的是代码能力,还有对栈的整体概念;有意思的是,这条题目的编码,相对而言其实会比之前一些中等难度的算法,还要难写,边界条件考虑的要很多;
思路
边界条件的考虑:考虑到存在这些括号字符的同时,也要考虑到不包含这些括号的字符串;
使用java
中stack类;
代码
package algorithm013;
import java.util.Stack;
public class Algorithm013 {
public static void main(String[] args) {
String s = "asdf";// false
String s2 = "()[]{}";// true
String s3 = "{(1)2}3[]4";// true
String s4 = "[}{()]";// false
String s5 = "[12(fgsf)4]";// true
String s6 = "{d[]df34}";// true
String s7 = "{f3[aer)}";// false
String s8 = "{3)32}";// false
System.out.println(isValidParentheses(s));
System.out.println(isValidParentheses(s2));
System.out.println(isValidParentheses(s3));
System.out.println(isValidParentheses(s4));
System.out.println(isValidParentheses(s5));
System.out.println(isValidParentheses(s6));
System.out.println(isValidParentheses(s7));
System.out.println(isValidParentheses(s8));
}
@SuppressWarnings("unchecked")
public static boolean isValidParentheses(String test) {
if (null != test && !"".equals(test)) {
@SuppressWarnings("rawtypes")
Stack stack = new Stack();
boolean isValid = false;
int length = test.length();
for (int i = 0; i < length; i++) {
char c = test.charAt(i);
if('(' == c || '{' == c || '[' == c ) {
isValid = true;
stack.push(c);
continue;
}
if(')' == c || '}' == c || ']' == c) {
isValid = true;
if(stack.isEmpty())
return false;
char pop = (char) stack.pop();
if(!(('(' == pop && ')' == c) || ('{' == pop && '}' == c) || ('[' == pop && ']' == c))) {
return false;
}
}
}
return isValid && stack.isEmpty();
}
return false;
}
}