From 1d206949dd3649f997868536b66b0d268e365965 Mon Sep 17 00:00:00 2001
From: lesleyzhao <tz2074@nyu.edu>
Date: Sun, 16 Jun 2024 13:27:11 +0800
Subject: [PATCH 1/2] =?UTF-8?q?=E6=9B=B4=E6=96=B0417=E4=B8=ADCarl=E9=A2=98?=
 =?UTF-8?q?=E8=A7=A3C++=E4=BB=A3=E7=A0=81=E9=87=8C=E7=9A=84=E6=B3=A8?=
 =?UTF-8?q?=E9=87=8A=E9=94=99=E8=AF=AF?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0417.太平洋大西洋水流问题.md | 6 +++---
 1 file changed, 3 insertions(+), 3 deletions(-)

diff --git a/problems/0417.太平洋大西洋水流问题.md b/problems/0417.太平洋大西洋水流问题.md
index b8448e93..5218b571 100644
--- a/problems/0417.太平洋大西洋水流问题.md
+++ b/problems/0417.太平洋大西洋水流问题.md
@@ -177,14 +177,14 @@ public:
 
         // 记录从大西洋出发，可以遍历的节点
         vector<vector<bool>> atlantic = vector<vector<bool>>(n, vector<bool>(m, false));
-
-        // 从最上最下行的节点出发，向高处遍历
+        
+        // 从最左最右列的节点出发，向高处遍历
         for (int i = 0; i < n; i++) {
             dfs (heights, pacific, i, 0); // 遍历最左列，接触太平洋 
             dfs (heights, atlantic, i, m - 1); // 遍历最右列，接触大西 
         }
 
-        // 从最左最右列的节点出发，向高处遍历
+        // 从最上最下行的节点出发，向高处遍历
         for (int j = 0; j < m; j++) {
             dfs (heights, pacific, 0, j); // 遍历最上行，接触太平洋
             dfs (heights, atlantic, n - 1, j); // 遍历最下行，接触大西洋

From 51f7960963061ad15bde0ff90067913d62b960ca Mon Sep 17 00:00:00 2001
From: lesleyzhao <tz2074@nyu.edu>
Date: Sun, 16 Jun 2024 13:50:09 +0800
Subject: [PATCH 2/2] =?UTF-8?q?=E6=9B=B4=E6=96=B0417Java=20DFS=E8=B7=9FCar?=
 =?UTF-8?q?l=20C++=E4=BB=A3=E7=A0=81=E6=80=9D=E8=B7=AF=E6=9B=B4=E4=B8=80?=
 =?UTF-8?q?=E8=87=B4=E7=9A=84=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 .../0417.太平洋大西洋水流问题.md    | 67 +++++++++++++++++++
 1 file changed, 67 insertions(+)

diff --git a/problems/0417.太平洋大西洋水流问题.md b/problems/0417.太平洋大西洋水流问题.md
index 5218b571..5156ce22 100644
--- a/problems/0417.太平洋大西洋水流问题.md
+++ b/problems/0417.太平洋大西洋水流问题.md
@@ -297,6 +297,73 @@ class Solution {
 }
 ```
 
+```Java
+class Solution {
+    
+    // 和Carl题解更加符合的Java DFS
+    private int[][] directions = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
+
+     /**
+     * @param heights 题目给定的二维数组
+     * @param m 当前位置的行号
+     * @param n 当前位置的列号
+     * @param visited 记录这个位置可以到哪条河
+     */
+
+    public void dfs(int[][] heights, boolean[][] visited, int m, int n){
+        if(visited[m][n]) return;        
+        visited[m][n] = true;
+
+        for(int[] dir: directions){
+            int nextm = m + dir[0];
+            int nextn = n + dir[1];
+            //出了2D array的边界，continue
+            if(nextm < 0||nextm == heights.length||nextn <0||nextn== heights[0].length) continue;
+            //下一个位置比当下位置还要低，跳过，继续找下一个更高的位置
+            if(heights[m][n] > heights[nextm][nextn]) continue;
+            dfs(heights, visited, nextm, nextn);
+        }
+    }
+
+
+    public List<List<Integer>> pacificAtlantic(int[][] heights) {
+        int m = heights.length;
+        int n = heights[0].length;
+
+        // 记录从太平洋边出发，可以遍历的节点
+        boolean[][] pacific = new boolean[m][n];
+        // 记录从大西洋出发，可以遍历的节点
+        boolean[][] atlantic = new boolean[m][n];
+        
+        // 从最左最右列的节点出发，向高处遍历
+        for(int i = 0; i<m; i++){
+            dfs(heights, pacific, i, 0); //遍历pacific最左边
+            dfs(heights, atlantic, i, n-1); //遍历atlantic最右边
+        }
+        
+        // 从最上最下行的节点出发，向高处遍历
+        for(int j = 0; j<n; j++){
+            dfs(heights, pacific, 0, j); //遍历pacific最上边
+            dfs(heights, atlantic, m-1, j); //遍历atlantic最下边
+        }
+
+        List<List<Integer>> result = new ArrayList<>();
+        for(int a = 0; a<m; a++){
+            for(int b = 0; b<n; b++){
+                // 如果这个节点，从太平洋和大西洋出发都遍历过，就是结果
+                if(pacific[a][b] && atlantic[a][b]){
+                   List<Integer> pair = new ArrayList<>();
+                   pair.add(a);
+                   pair.add(b);
+                   result.add(pair);
+                }
+            }
+        }
+        return result;
+    }
+}
+```
+
 广度优先遍历：
 
 ```Java