单调栈结构(进阶) + 栈的使用 + 思维

单调栈结构(进阶)

单调栈结构(进阶)

题目描述

方法一：暴力法

@Test
    public void main1(){
        Scanner cin = new Scanner(System.in);
        int n = cin.nextInt();
        int[] nums = new int[n];
        for(int i=0; i<n; i++){
            nums[i] = cin.nextInt();
        }
        int mins = 1000006;
        int mini = -1;
        int[] min = new int[n];
        int[] max = new int[n];
        for(int i=0; i<n; i++){
            int j;
            for(j=i-1; j>=0; j--){
                if(nums[j] < nums[i]){
                    min[i] = j;
                    break;
                }
            }
            if(j == -1)
                min[i] = -1;
            for(j=i+1; j<n; j++){
                if(nums[j] < nums[i]){
                    max[i] = j;
                    break;
                }
            }
            if(j == n)
                max[i] = -1;
        }
        for(int i=0; i<n; i++){
            System.out.println(min[i] + " " + max[i]);
        }
    }

方法二：使用单调递增栈维护单向的最小值

public static void main(String[] args){
        Scanner cin = new Scanner(System.in);
        int n = cin.nextInt();
        int[] nums = new int[n];
        for(int i=0; i<n; i++){
            nums[i] = cin.nextInt();
        }
        Stack<Integer> sta = new Stack<>();//严格递增栈，存储序号
        int[] mins = new int[n];
        int[] maxs = new int[n];
        for(int i=0; i<n; i++){
            while(!sta.isEmpty() && nums[sta.peek()] >= nums[i]){
                sta.pop();
            }
            if(!sta.isEmpty()){
                mins[i] = sta.peek();
            }else mins[i] = -1;
            sta.push(i);
        }
        sta.clear();
        for(int i=n-1; i>=0; i--){
            while(!sta.isEmpty() && nums[sta.peek()] >= nums[i]){
                sta.pop();
            }
            if(!sta.isEmpty()){
                maxs[i] = sta.peek();
            }else maxs[i] = -1;
            sta.push(i);
        }
        for(int i=0; i<n; i++){
            System.out.println(mins[i] + " " + maxs[i]);
        }

    }