Day 11 | 20. Valid Parentheses | 1047. Remove All Adjacent Duplicates In String | 150. Evaluate RPN

LeetCode实战：从搜索到数据结构，一周算法之旅

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博主分享了一周内涵盖多种LeetCode题目，涉及二分查找、栈与队列、字符串操作、数组问题、图算法等，包括Valid Parentheses的括号验证，Remove Adjacent Duplicates的字符串处理，以及Reverse Polish Notation的计算表达式。通过实例展示了算法解决思路和数据结构的应用。

LeetCode 20. Valid Parentheses

Question Link

Solution:

class Solution {
    public boolean isValid(String s) {
        // if the length of s is an odd number, s must be invalid
        if(s.length()%2 != 0) return false;

        Stack<Character> stack   = new Stack<>();
        for(int i = 0; i < s.length(); i++){
            char ch = s.charAt(i);
            if(ch == '(' )
                stack.push(')');
            else if(ch == '[')
                stack.push(']');
            else if(ch == '{')
                stack.push('}');
            // First check if the stack is empty, then judge whether the top element is equal to the current letter
            // Situation 3 and 2
            else if(stack.isEmpty() || stack.peek() != ch)
                return false;
            else
                stack.pop();    
        }
        //Int 1 situation, the stack is not empty
        return stack.isEmpty();
    }
}

Thought:

We need to deal with three situations:
- ① ( [ { } ] ( )
- ② [ { ( ] } ]
- ③ [ { } ] ( ) ) )

LeetCode 1047. Remove All Adjacent Duplicates In String

Question Link

Solution:

class Solution {
    public String removeDuplicates(String s) {
        Stack<Character> stack = new Stack();
        for(int i = 0; i < s.length(); i++){
            char ch = s.charAt(i);
            if(stack.isEmpty() || stack.peek()!=ch)
                stack.push(ch);
            else
                stack.pop();
        }
        // The remaining elements in the stack are unrepeated.
        String str = "";
        while(!stack.isEmpty())
            str = stack.pop() + str;
        return str.toString();
    }
}

Thought:

Use a stack to save traversed elements. When traversing the current element, check the stack to see whether there is an adjacent duplicated element.
If exists repeated element, perform elimination operation pop(). Otherwise do push()
The remaining elements in the stack are unrepeated.

LeetCode 150. Evaluate Reverse Polish Notation

Question Link

Solution:

class Solution {
    public int evalRPN(String[] tokens) {
        Deque<Integer> stack = new LinkedList<>();
        for(String s : tokens){
            if(s.equals("+"))
                stack.push(stack.pop() + stack.pop());
            else if(s.equals("-"))  
                stack.push(-stack.pop() + stack.pop());
            else if(s.equals("*"))
                stack.push(stack.pop() * stack.pop());
            else if(s.equals("/")){
                int temp1 = stack.pop();
                int temp2 = stack.pop();
                stack.push(temp2/temp1);
            }
            else
                stack.push(Integer.valueOf(s));
        }
        return stack.pop();
    }
}

Thoughts:

Reverse Polish Notation is a suffix notation. The so-called suffix means the indicator is in the back.
For example: This RPE of ( 1 + 2 ) * ( 3 + 4 ) is written as ( ( 1 2 + ) ( 3 4 + ) * )
Use the stack to operate RPN calculation. If it encounters a number, do push(). If it meets an indicator, get the numbers in the stack to do the calculation. Then push the result to the stack.
Note: - and / require additional processing.