我的面試準(zhǔn)備過程--隊列與棧(更新中)

EastWoodYang 發(fā)布于2019-08-16 10:38 / 839人閱讀

摘要：和三個方法的時間復(fù)雜度必須為兩種解法，解法一，將最小值存入自有的數(shù)據(jù)結(jié)構(gòu)中，如下所示原本的值最小值解法二，用兩個棧

堆棧和隊列統(tǒng)稱線性表

簡單的線性結(jié)構(gòu)

數(shù)組和鏈表可以實現(xiàn)這兩種數(shù)據(jù)結(jié)構(gòu)

堆棧

基本理解

DFS

深度優(yōu)先---按深度遍歷

遞歸轉(zhuǎn)非遞歸

隊列

基本理解

BFS

廣度優(yōu)先---按層序遍歷

出入棧的合法性
模擬出入棧的過程，不是入棧，就是出棧，不然就不合法

public boolean isPossible(int[] in, int[] out){
    if(in.length != out.length){
        return false;
    }
    
    Stack s = new Stack<>();
    for(int i = 0, j = 0;j < out.length; j++){
        //如果不是入棧的
        while(s.isEmpty() && s.peek() != out[j]){
            if(i >= in.length){
                return false;
            }
            s.push(in[i++]);
        }
        //那就出棧
        s.pop();
    }
    
    return true;
}

兩個棧實現(xiàn)隊列

    public class MyQueue{
        Stack s1 = new Stack<>();
        Stack s2 = new Stack<>();

        public void push(int x){
            s1.push(x);
        }

        //負(fù)負(fù)得正
        public int pop(){
            if(s2.isEmpty()){
                while(!s1.isEmpty()){
                    s2.push(s1.pop());
                }
            }
            return s2.pop();
        }

        public int peek(){
            if(s2.isEmpty()){
                while(!s1.isEmpty()){
                    s2.push(s1.pop());
                }
            }
            return s2.peek();
        }

        public boolean empty(){
            return s1.isEmpty() && s2.isEmpty();
        }

    }

兩個隊列實現(xiàn)棧

public class MyStack {
    Queue queue1;
    Queue queue2;
    /** Initialize your data structure here. */
    public MyStack() {
        queue1 = new LinkedList<>();
        queue2 = new LinkedList<>();
    }
    
    /** Push element x onto stack. */
    public void push(int x) {
        if(!queue1.isEmpty()){
            queue1.offer(x);
        }else{
            queue2.offer(x);
        }
    }
    
    /** Removes the element on top of the stack and returns that element. */
    public int pop() {
        if(queue1.size() != 0){
            while(queue1.size() > 1){
                queue2.add(queue1.peek());
                queue1.poll();
            }
            return queue1.remove();
        }else{
            while(queue2.size() > 1){
                queue1.add(queue2.peek());
                queue2.poll();
            }
            return queue2.remove();
        }
    }
    
    /** Get the top element. */
    public int top() {
        if(queue1.size() != 0){
            while(queue1.size() > 1){
                queue2.add(queue1.poll());
            }
            int res = queue1.peek();
            queue2.add(queue1.poll());
            return res;
        }else{
            while(queue2.size() > 1){
                queue1.add(queue2.poll());
            }
            int res = queue2.peek();
            queue1.add(queue2.poll());
            return res;
        }
    }
    
    /** Returns whether the stack is empty. */
    public boolean empty() {
        return queue1.isEmpty() && queue2.isEmpty();
    }
}

設(shè)計一個棧，除pop與push方法，還支持min方法，可返回棧元素中的最小值。push、pop和min三個方法的時間復(fù)雜度必須為O(1)

兩種解法，解法一，將最小值存入自有的數(shù)據(jù)結(jié)構(gòu)中，如下所示

public class MyStack extends Stack{
    
    public void push(int value){
        int newMin = Math.min(value, min());
        super.push(new NodeWithMin(value, newMin));
    }
    
    public int min(){
        if(this.isEmpty()){
            return Integer.MAX_VALUE;
        }else{
            return peek().min;
        }
    }
    
    public int pop(){
        super.pop().value;
    }
}

class NodeWithMin{
    int value;//原本的值
    int min;//最小值
    public NodeWithMin(int value, int min){
        this.value = value;
        this.min = min;
    }
}

解法二，用兩個棧

public class MyStack{
        Stack s1 = new Stack<>();
        Stack s2 = new Stack<>();

        public void push(int i){
                s1.push(i);
                if(s2.isEmpty() || min() >= i){
                        s2.push(i);
                }
        }

        public int pop(){
                if(s1.peek() == min()){
                        s2.pop();
                }

                return s1.pop();
        }

        public int min(){
                return s2.peek();
        }

}