1.删除字符串中的所有相邻重复项

可以用数组模拟栈结构 

class Solution {
    public String removeDuplicates(String s) {
        if(s.length()<=1){
            return s;
        }
        StringBuffer ret=new StringBuffer();
        
        for(int i=0;i<s.length();i++){
            if(ret.length()<1){
                ret.append(s.charAt(i));
            }else if(s.charAt(i)==ret.charAt(ret.length()-1)){
                ret.deleteCharAt(ret.length()-1);
            }else{
                ret.append(s.charAt(i));
            }
        }
        return ret.toString();
    }
}

 

2.比较含退格的字符串 

844. 比较含退格的字符串 - 力扣（LeetCode）

class Solution {
    public boolean backspaceCompare(String s, String t) {
        return changeStr(s).equals(changeStr(t));
    }

    public String changeStr(String str1){
        StringBuffer ss=new StringBuffer();
        char[] str=str1.toCharArray();
        for(char ch:str){
            if(ch!='#'){
                ss.append(ch);//入栈
            }else{
                if(ss.length()>0){
                    ss.deleteCharAt(ss.length()-1);//出栈
                }
            }
        }
        return ss.toString();
    }
}

3.基本计算器2

. - 力扣（LeetCode）

算法原理

双端队列也可以实现栈

class Solution {
    public int calculate(String s) {
        //双端队列，实现栈操作
        Deque<Integer> deque=new ArrayDeque<>();
        char op='+';
        int i=0,n=s.length();
        char[] ss=s.toCharArray();
        while(i<n){
            //1.先考虑为空的情况
            if(ss[i]==' '){
                i++;
            }else if(ss[i]>='0'&ss[i]<='9'){
                int tmp=0;
                while(i<n&&ss[i]>='0'&ss[i]<='9'){
                    tmp=tmp*10+(ss[i]-'0');
                    i++;
                }
                if(op=='+'){
                    deque.push(tmp);
                }else if(op=='-'){
                    deque.push(-tmp);
                }else if(op=='*'){
                    deque.push(deque.poll()*tmp);
                }else if(op=='/'){
                    deque.push(deque.poll()/tmp);
                }
            }else{
                op=ss[i];
                i++;
            }
        }
        //遍历队列
        int ret=0;
        while(!deque.isEmpty()){
            ret=ret+deque.poll();
        }
        return ret;
    }
}

 4.字符串解码

394. 字符串解码 - 力扣（LeetCode）

 算法原理

class Solution {
    public String decodeString(String s) {
        int i=0,n=s.length();
        char[] ss=s.toCharArray();
        //构造栈
        Stack<StringBuffer> strstack=new Stack<>();
        //字符串要先加入一个空字符串
        //不然会出现越界的情况
        strstack.push(new StringBuffer());
        Stack<Integer> numstack=new Stack<>();
        while(i<n){
            //1.遇到数字的情况
            if(ss[i]>='0'&&ss[i]<='9'){
                int tmp=0;
                while(i<n&&ss[i]>='0'&&ss[i]<='9'){
                    tmp=tmp*10+(ss[i]-'0');
                    i++;
                }
                numstack.push(tmp);
            }else if(ss[i]==' '){
                i++;
            }else if(ss[i]=='['){
                i++;
                //后面遇到的字符串都要加入栈中
                StringBuffer str=new StringBuffer();
                while(i<n&&ss[i]>='a'&&ss[i]<='z'){
                    str.append(ss[i]);
                    i++;
                }
                strstack.push(str);
            }else if(ss[i]==']'){
                int num1=numstack.pop();
                StringBuffer str1=strstack.pop();
                while(num1--!=0){
                    strstack.peek().append(str1);
                }
                i++;
            }else{
                //后面遇到的字符串都要加入栈中
                StringBuffer str=new StringBuffer();
                while(i<n&&ss[i]>='a'&&ss[i]<='z'){
                    str.append(ss[i]);
                    i++;
                }
                strstack.peek().append(str);
            }
        }
        return strstack.peek().toString();
    }
}

 5.验证栈序列

. - 力扣（LeetCode）

class Solution {
    public boolean validateStackSequences(int[] pushed, int[] popped) {
        Stack<Integer> st = new Stack<>();
        int i = 0, n = popped.length;
        for (int x : pushed) {
            st.push(x);
            while (!st.isEmpty() && st.peek() == popped[i]) {
                st.pop();
                i++;
            }
        }
        return i == n;
    }
}