力扣-图论-17【算法学习day.67】

前言

###我做这类文章一个重要的目的还是给正在学习的大家提供方向和记录学习过程（例如想要掌握基础用法，该刷哪些题？）我的解析也不会做的非常详细，只会提供思路和一些关键点，力扣上的大佬们的题解质量是非常非常高滴！！！

习题

1.检查网格中是否存在有效路径

题目链接:1391. 检查网格中是否存在有效路径 - 力扣（LeetCode）分析:就是dfs，判断条件比较繁琐 

代码:

class Solution {
    int[][] grid;
    int n,m;
    int[][] flag;
    boolean ans = false;
    public boolean hasValidPath(int[][] grid) {
        this.grid = grid;
        n = grid.length;
        m = grid[0].length;
        flag = new int[n][m];
        flag[0][0] = 1;
        recursion(0,0);
        return ans;
    }
    public void recursion(int x,int y){
        // System.out.println(x+"          "+y);
        if(x==n-1&&y==m-1){
           ans = true;
           return;
        }
        if(x+1<n&&flag[x+1][y]==0){
             flag[x+1][y] = 1;
             if(grid[x][y]==2&&grid[x+1][y]==2){
                recursion(x+1,y);
             }
            if(grid[x][y]==2&&(grid[x+1][y]==5||grid[x+1][y]==6)){
                recursion(x+1,y);
            }
            if((grid[x][y]==3||grid[x][y]==4)&&grid[x+1][y]==2){
                recursion(x+1,y);
            }
            if((grid[x][y]==3||grid[x][y]==4)&&(grid[x+1][y]==5||grid[x+1][y]==6)){
                recursion(x+1,y);
            }
            flag[x+1][y] = 0;
        }
        if(x-1>=0&&flag[x-1][y]==0){
             flag[x-1][y] = 1;
              if(grid[x][y]==2&&grid[x-1][y]==2){
                recursion(x-1,y);
             }
            if((grid[x][y]==5||grid[x][y]==6)&&(grid[x-1][y]==3||grid[x-1][y]==4)){
                recursion(x-1,y);
            }
            if((grid[x][y]==5||grid[x][y]==6)&&grid[x-1][y]==2){
                recursion(x-1,y);
            }
            if(grid[x][y]==2&&(grid[x-1][y]==3||grid[x-1][y]==4)){
                recursion(x-1,y);
            }
             flag[x-1][y] = 0;
        }
        if(y+1<m&&flag[x][y+1]==0){
            flag[x][y+1] = 1;
             if(grid[x][y]==1&&grid[x][y+1]==1){
                recursion(x,y+1);
             }
            if(grid[x][y]==1&&(grid[x][y+1]==3||grid[x][y+1]==5)){
                recursion(x,y+1);
            }
            if((grid[x][y]==4||grid[x][y]==6)&&grid[x][y+1]==1){
                recursion(x,y+1);
            }
            if((grid[x][y]==4||grid[x][y]==6)&&(grid[x][y+1]==3||grid[x][y+1]==5)){
                recursion(x,y+1);
            }
            flag[x][y+1] = 0;
        }
        if(y-1>=0&&flag[x][y-1]==0){
            flag[x][y-1]=1;
            if(grid[x][y]==1&&grid[x][y-1]==1){
                recursion(x,y-1);
             }
            if(grid[x][y]==1&&(grid[x][y-1]==4||grid[x][y-1]==6)){
                recursion(x,y-1);
            }
            if((grid[x][y]==3||grid[x][y]==5)&&grid[x][y-1]==1){
                recursion(x,y-1);
            }
            if((grid[x][y]==3||grid[x][y]==5)&&(grid[x][y-1]==4||grid[x][y-1]==6)){
                recursion(x,y-1);
            }
            flag[x][y-1] = 0;
        }
    }
}

2.太平洋大西洋水流问题

题目链接:417. 太平洋大西洋水流问题 - 力扣（LeetCode） 

题面:

分析:dfs，用两个变量来判断是否既能流入太平洋又能流入大西洋，复杂度较高 

代码: 

class Solution {
    int[][] heights;
    int n,m;
    int[][] flag;
    int flag2 = 0;
    int flag3 = 0;
    List<Integer> list;
    List<List<Integer>> ans = new ArrayList<>();
    public List<List<Integer>> pacificAtlantic(int[][] heights) {
        this.heights = heights;
        n = heights.length;
        m = heights[0].length;
        for(int i = 0;i<n;i++){
            for(int j = 0;j<m;j++){
                 flag2 = 0;
                 flag3 = 0;
                 list = new ArrayList<>();
                 flag = new int[n][m];
                 flag[i][j] = 1;
                 recursion(i,j);
                 if(flag2==1&&flag3==1){
                    list.add(i);
                    list.add(j);
                    ans.add(list);
                 }
            }
        }
        return ans;
    }
    public void recursion(int x,int y){
        if(x==0||y==0)flag2 = 1;
        if(x==n-1||y==m-1)flag3 = 1;
        if(flag2==1&&flag3==1)return;
        if(x+1<n&&heights[x+1][y]<=heights[x][y]&&flag[x+1][y]==0){
            flag[x+1][y] = 1;
            recursion(x+1,y);
            flag[x+1][y] = 0;
        }
        if(x-1>=0&&heights[x-1][y]<=heights[x][y]&&flag[x-1][y]==0){
            flag[x-1][y] = 1;
            recursion(x-1,y);
            flag[x-1][y] = 0;
        }
        if(y+1<m&&heights[x][y+1]<=heights[x][y]&&flag[x][y+1]==0){
            flag[x][y+1] = 1;
            recursion(x,y+1);
            flag[x][y+1] = 0;
        }
        if(y-1>=0&&heights[x][y-1]<=heights[x][y]&&flag[x][y-1]==0){
            flag[x][y-1] = 1;
            recursion(x,y-1);
            flag[x][y-1] = 0;
        }
    }
}

 后言

上面是力扣图论专题，下一篇是其他的习题，希望有所帮助，一同进步，共勉！