一、题目
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. Example 1: Input: "()" Output: true Example 2: Input: "()[]{}" Output: true Example 3: Input: "(]" Output: false Example 4: Input: "([)]" Output: false Example 5: Input: "{[]}" Output: true
二、解题思路
1、利用List集合实现一个栈; 2、将字符串s转换成字符数组,并循环遍历; 3、如果字符为:"{、(、["中的一个,则存入集合中; 4、如果字符为:"}、)、]"中的一个,则取出集合中最后一个元素进行比较; 5、如能匹配上,则删除集合中最后一个元素,否则返回false; 6、最后判断集合大小是否为0,如是则返回true。
三、代码实现
public boolean isValid(String s) {
if ("".equals(s)) {
return true;
} else {
Map<Character, Character> parentheseMap = new HashMap<Character, Character>();
parentheseMap.put(')', '(');
parentheseMap.put(']', '[');
parentheseMap.put('}', '{');
char[] sArr = s.toCharArray();
List<Character> stackList = new ArrayList<Character>();
for (int i = 0; i < sArr.length; i++) {
if (sArr[i] == '(' || sArr[i] == '[' || sArr[i] == '{') {
stackList.add(sArr[i]);
} else {
if (stackList.size() == 0) {
return false;
} else {
char temp = stackList.get(stackList.size() - 1);
if (temp == parentheseMap.get(sArr[i])) {
stackList.remove(stackList.size() - 1);
} else {
return false;
}
}
}
}
return stackList.size() == 0 ? true : false;
}
}