day11算法补卡|栈与队列02|Leetcode20有效括号、1047删除字符串中的所有相邻重复项 、150逆波兰表达式求值

Leetcode20:有效括号

题目链接:https://leetcode.cn/problems/valid-parentheses/description/

 题目分析：使用栈实现，如果栈为空，直接入栈；如果栈不为空且栈顶元素与即将遍历字符匹配，则对应栈顶元素出栈；如不匹配，则对元素直接入栈

Java实现代码:

import java.util.Stack;

//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
    public boolean isValid(String s) {
        int length=s.length();
        Stack strstack = new Stack<>();
        for (int i = 0; i < length; i++) {
            char thechar = s.charAt(i);
            if(strstack.isEmpty()){
                strstack.push(thechar);
                continue;
            }
            char stacktop = strstack.peek();
            if((thechar == ')' && stacktop == '(')||(thechar == '}' && stacktop == '{')||(thechar == ']' && stacktop == '[')){
                strstack.pop();
            }else {
                strstack.push(thechar);
            }
        }
        return strstack.isEmpty();
    }
}
//leetcode submit region end(Prohibit modification and deletion)

Leetcode1047 .删除字符串中的所有相邻重复项 

题目链接：https://leetcode.cn/problems/remove-all-adjacent-duplicates-in-string/description/

题目分析：与括号匹配思路一致 均属于消消消问题

Java实现代码：

class Solution {
    public String removeDuplicates(String s) {
        int leng = s.length();
        Stack strstack = new Stack<>();
        for (int i = 0; i < leng; i++) {
            char thechar = s.charAt(i);
            if(strstack.isEmpty()){
                strstack.push(thechar);
                continue;
            }
            if(strstack.peek() == thechar){
                strstack.pop();
            }else {
                strstack.push(thechar);
            }
        }
        String returnstr = "";
        while (!strstack.isEmpty()){
            returnstr = strstack.pop()+returnstr;
        }

        return returnstr;
    }
}

Leetcode150.逆波兰表达式求值

题目链接：https://leetcode.cn/problems/evaluate-reverse-polish-notation/description/

主要是'逆向'--要注意其中的除法与减法计算

主要思路是：遇到符号---两次pop 运算后进行依次push;不是符号直接push

Java实现代码：

import java.util.Stack;

class Solution {
    public int evalRPN(String[] tokens) {
        Stack strstack = new Stack<>();
        int size = tokens.length;
        for (int i =0;i

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今日补卡结束:对于栈与队列相关问题，主要是把握其'消消消'特性与顺序问题，想清楚能够使用时就很好写出题解；