From 30d4b36c27b1c9293f333a243d3075a687e4e2ff Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <tangshusen@pku.edu.cn>
Date: Wed, 17 Jul 2019 10:25:12 +0800
Subject: [PATCH] Create 130. Surrounded Regions.md

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+# [130. Surrounded Regions](https://leetcode.com/problems/surrounded-regions/)
+# 思路
+有一个二维的只包含O和X的矩形区域，题目要求将被X包住的O都变成X，边缘的O不算被包围。            
+这题很容易想到的是遍历一遍矩形，若遇到了O，就从这个O出发进行DFS（或者BFS），看会不会遇到边界，如果遇不到则说明是被包围的，然后将这些O变成X。
+
+但是上面的思路在矩形很大的时候亲测会超时，我们可以用反向思路考虑: 
+扫描矩阵的四条边，如果有O，则用DFS遍历，将所有连着的O都变成另一个字符比如N，这样剩下的O都是被包围的，最后将这些O变成X，把N变回O就行了。
+由于这个思路是只从矩形的四条边进行DFS，所以比最开始的思路复杂度要低很多（虽然理论复杂度都是O(m*n), m和n分别是矩形的高和宽）。
+
+
+# C++
+``` C++
+class Solution {
+private:
+    vector<int>dx{1, -1, 0, 0};
+    vector<int>dy{0, 0, 1, -1};
+    int m, n;
+    void DFS(vector<vector<char>>& board, int i, int j){
+        if(i < 0 || j < 0 || i > m-1 || j > n-1) return; // 超出矩形区域
+        if(board[i][j] == 'X' || board[i][j] == 'N') return;
+        
+        // board[i][j] == 'O'
+        board[i][j] = 'N';
+        int next_i, next_j;
+        for(int k = 0; k < 4; k++) {
+            next_i = i+dx[k];
+            next_j = j+dy[k];
+            // 提前判断一下会快不少
+            if(next_i < 0 || next_j < 0 || next_i > m-1 || next_j > n-1) continue;
+            DFS(board, next_i, next_j);
+        }      
+    }
+public:
+    void solve(vector<vector<char>>& board) {
+        m = board.size();
+        if(m == 0) return;
+        n = board[0].size();
+        
+        for(int j = 0; j < n; j++){
+            DFS(board, 0, j);
+            DFS(board, m-1, j);
+        }
+        
+        for(int i = 1; i < m-1; i++){
+            DFS(board, i, 0);
+            DFS(board, i, n-1);
+        }
+                          
+        for(int i = 0; i < m; i++)
+            for(int j = 0; j < n; j++){
+                if(board[i][j] == 'N') board[i][j] = 'O';
+                else if(board[i][j] == 'O') board[i][j] = 'X';
+            }
+    }
+};
+```