一、栈和队列

1.设计一个有getMin功能的栈

定义栈的数据结构，请在该类型中实现一个能够得到栈中所含最小元素的min函数（时间复杂度应为O（1））。

import java.util.Stack;class Main {    Stack
     
       dataStack = new Stack<>();    Stack
      
        minStack = new Stack<>();    public static void main(String[] args) {        Main stack = new Main();        stack.push(11);        stack.push(2);        stack.pop();        stack.push(70);        stack.push(51);        System.out.println(stack.getMin());        stack.push(4);        stack.push(113);        System.out.println(stack.getMin());    }    public void push(int num) {        if(minStack.isEmpty())             minStack.push(num);        if(num > minStack.peek()) {            minStack.push(minStack.peek());        } else {            minStack.push(num);        }        dataStack.push(num);    }    public int pop() {        if(dataStack.isEmpty()) {            throw new RuntimeException("Your stack is empty.");        }        minStack.pop();        return dataStack.pop();    }    public int getMin() {        if(minStack.isEmpty()) {            throw new RuntimeException("Your stack is empty.");        }        return minStack.peek();    }}
      
     

 

class Solution {    public int numIslands(char[][] grid) {                if(grid == null || grid.length <= 0 || grid[0].length <= 0){            return 0;        }        int ans = 0;        int n = grid.length;        int m = grid[0].length;        for(int i=0; i
      
       =n || y>=m || mat[x][y] != '1') return;        mat[x][y]='2';        dfs(mat,x+1,y,n,m);        dfs(mat,x-1,y,n,m);        dfs(mat,x,y+1,n,m);        dfs(mat,x,y-1,n,m);    }}
      

 

class Solution {    public int maxAreaOfIsland(int[][] grid) {        if(grid==null || grid.length==0 || grid[0].length==0) return 0;        int n = grid.length;        int m =  grid[0].length;        int ans = 0;        for(int i=0; i
      
       =n || y>=m || a[x][y] != 1) return 0;        a[x][y] = 2;        ans += dfs(a,x+1,y,n,m);        ans += dfs(a,x-1,y,n,m);        ans += dfs(a,x,y+1,n,m);        ans += dfs(a,x,y-1,n,m);        return ans;     }}