南柯一梦

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<Integer> dataStack = new Stack<>();
    Stack<Integer> 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; i++){
            for(int j=0; j<m; j++){
                if(grid[i][j] == '1'){
                    ans++;
                    dfs(grid,i,j,n,m);
                }
            }
        }
        return ans;
    }
    public void dfs(char[][] mat, int x, int y, int n, int m){
        if(x<0 || y<0 || x>=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; i++)
            for(int j=0; j<m; j++){
                if(grid[i][j] == 1){
                    int cnt = dfs(grid,i,j,n,m);
                    ans = Math.max(ans,cnt);
                }
            }
        return ans;
    }
    
    public int dfs(int[][] a, int x, int y, int n, int m){
        int ans = 1;
        if(x<0 || y<0 || x>=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;
     }
}