From 836191f40850618cd526b16ceba98f8f78190c9e Mon Sep 17 00:00:00 2001
From: CyC2018 <zhengyc101@163.com>
Date: Fri, 3 May 2019 19:22:28 +0800
Subject: [PATCH] auto commit

---
 docs/notes/剑指 Offer 题解 - 10~19.md         |  30 ++++++++++++------
 .../dc964b86-7a08-4bde-a3d9-e6ddceb29f98.png  | Bin 0 -> 11864 bytes
 .../dc964b86-7a08-4bde-a3d9-e6ddceb29f98.png  | Bin 0 -> 11864 bytes
 notes/剑指 Offer 题解 - 10~19.md              |  30 ++++++++++++------
 4 files changed, 40 insertions(+), 20 deletions(-)
 create mode 100644 docs/pics/dc964b86-7a08-4bde-a3d9-e6ddceb29f98.png
 create mode 100644 notes/pics/dc964b86-7a08-4bde-a3d9-e6ddceb29f98.png

diff --git a/docs/notes/剑指 Offer 题解 - 10~19.md b/docs/notes/剑指 Offer 题解 - 10~19.md
index e4e4ec26..d31c19ed 100644
--- a/docs/notes/剑指 Offer 题解 - 10~19.md	
+++ b/docs/notes/剑指 Offer 题解 - 10~19.md	
@@ -296,7 +296,7 @@ private int minNumber(int[] nums, int l, int h) {
 
 ## 题目描述
 
-请设计一个函数，用来判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一个格子开始，每一步可以在矩阵中向左，向右，向上，向下移动一个格子。如果一条路径经过了矩阵中的某一个格子，则该路径不能再进入该格子。
+判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一个格子开始，每一步可以在矩阵中向左，向右，向上，向下移动一个格子。如果一条路径经过了矩阵中的某一个格子，则该路径不能再进入该格子。
 
 例如下面的矩阵包含了一条 bfce 路径。
 
@@ -304,14 +304,19 @@ private int minNumber(int[] nums, int l, int h) {
 
 ## 解题思路
 
+使用回溯法（backtracking）进行求解，它是一种暴力搜索方法，通过搜索所有可能的结果来求解问题。回溯法在一次搜索结束时需要进行回溯（回退），将这一次搜索过程中设置的状态进行清除，从而开始一次新的搜索过程。例如下图示例中，从 f 开始，下一步有 4 种搜索可能，如果先搜索 b，需要将 b 标记为已经使用，防止重复使用。在这一次搜索结束之后，需要将 b 的已经使用状态清除，并搜索 c。
+
+<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/dc964b86-7a08-4bde-a3d9-e6ddceb29f98.png" width="200px"> </div><br>
+
+本题的输入是数组而不是矩阵（二维数组），因此需要先将数组转换成矩阵。
+
 ```java
 private final static int[][] next = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};
 private int rows;
 private int cols;
 
 public boolean hasPath(char[] array, int rows, int cols, char[] str) {
-    if (rows == 0 || cols == 0)
-        return false;
+    if (rows == 0 || cols == 0) return false;
     this.rows = rows;
     this.cols = cols;
     boolean[][] marked = new boolean[rows][cols];
@@ -320,14 +325,19 @@ public boolean hasPath(char[] array, int rows, int cols, char[] str) {
         for (int j = 0; j < cols; j++)
             if (backtracking(matrix, str, marked, 0, i, j))
                 return true;
+
     return false;
 }
 
-private boolean backtracking(char[][] matrix, char[] str, boolean[][] marked, int pathLen, int r, int c) {
-    if (pathLen == str.length)
-        return true;
-    if (r < 0 || r >= rows || c < 0 || c >= cols || matrix[r][c] != str[pathLen] || marked[r][c])
+private boolean backtracking(char[][] matrix, char[] str,
+                             boolean[][] marked, int pathLen, int r, int c) {
+
+    if (pathLen == str.length) return true;
+    if (r < 0 || r >= rows || c < 0 || c >= cols
+            || matrix[r][c] != str[pathLen] || marked[r][c]) {
+
         return false;
+    }
     marked[r][c] = true;
     for (int[] n : next)
         if (backtracking(matrix, str, marked, pathLen + 1, r + n[0], c + n[1]))
@@ -338,9 +348,9 @@ private boolean backtracking(char[][] matrix, char[] str, boolean[][] marked, in
 
 private char[][] buildMatrix(char[] array) {
     char[][] matrix = new char[rows][cols];
-    for (int i = 0, idx = 0; i < rows; i++)
-        for (int j = 0; j < cols; j++)
-            matrix[i][j] = array[idx++];
+    for (int r = 0, idx = 0; r < rows; r++)
+        for (int c = 0; c < cols; c++)
+            matrix[r][c] = array[idx++];
     return matrix;
 }
 ```
diff --git a/docs/pics/dc964b86-7a08-4bde-a3d9-e6ddceb29f98.png b/docs/pics/dc964b86-7a08-4bde-a3d9-e6ddceb29f98.png
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HcmV?d00001

diff --git a/notes/剑指 Offer 题解 - 10~19.md b/notes/剑指 Offer 题解 - 10~19.md
index 6902dddd..8bb22cd0 100644
--- a/notes/剑指 Offer 题解 - 10~19.md	
+++ b/notes/剑指 Offer 题解 - 10~19.md	
@@ -296,7 +296,7 @@ private int minNumber(int[] nums, int l, int h) {
 
 ## 题目描述
 
-请设计一个函数，用来判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一个格子开始，每一步可以在矩阵中向左，向右，向上，向下移动一个格子。如果一条路径经过了矩阵中的某一个格子，则该路径不能再进入该格子。
+判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一个格子开始，每一步可以在矩阵中向左，向右，向上，向下移动一个格子。如果一条路径经过了矩阵中的某一个格子，则该路径不能再进入该格子。
 
 例如下面的矩阵包含了一条 bfce 路径。
 
@@ -304,14 +304,19 @@ private int minNumber(int[] nums, int l, int h) {
 
 ## 解题思路
 
+使用回溯法（backtracking）进行求解，它是一种暴力搜索方法，通过搜索所有可能的结果来求解问题。回溯法在一次搜索结束时需要进行回溯（回退），将这一次搜索过程中设置的状态进行清除，从而开始一次新的搜索过程。例如下图示例中，从 f 开始，下一步有 4 种搜索可能，如果先搜索 b，需要将 b 标记为已经使用，防止重复使用。在这一次搜索结束之后，需要将 b 的已经使用状态清除，并搜索 c。
+
+<div align="center"> <img src="pics/dc964b86-7a08-4bde-a3d9-e6ddceb29f98.png" width="200px"> </div><br>
+
+本题的输入是数组而不是矩阵（二维数组），因此需要先将数组转换成矩阵。
+
 ```java
 private final static int[][] next = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};
 private int rows;
 private int cols;
 
 public boolean hasPath(char[] array, int rows, int cols, char[] str) {
-    if (rows == 0 || cols == 0)
-        return false;
+    if (rows == 0 || cols == 0) return false;
     this.rows = rows;
     this.cols = cols;
     boolean[][] marked = new boolean[rows][cols];
@@ -320,14 +325,19 @@ public boolean hasPath(char[] array, int rows, int cols, char[] str) {
         for (int j = 0; j < cols; j++)
             if (backtracking(matrix, str, marked, 0, i, j))
                 return true;
+
     return false;
 }
 
-private boolean backtracking(char[][] matrix, char[] str, boolean[][] marked, int pathLen, int r, int c) {
-    if (pathLen == str.length)
-        return true;
-    if (r < 0 || r >= rows || c < 0 || c >= cols || matrix[r][c] != str[pathLen] || marked[r][c])
+private boolean backtracking(char[][] matrix, char[] str,
+                             boolean[][] marked, int pathLen, int r, int c) {
+
+    if (pathLen == str.length) return true;
+    if (r < 0 || r >= rows || c < 0 || c >= cols
+            || matrix[r][c] != str[pathLen] || marked[r][c]) {
+
         return false;
+    }
     marked[r][c] = true;
     for (int[] n : next)
         if (backtracking(matrix, str, marked, pathLen + 1, r + n[0], c + n[1]))
@@ -338,9 +348,9 @@ private boolean backtracking(char[][] matrix, char[] str, boolean[][] marked, in
 
 private char[][] buildMatrix(char[] array) {
     char[][] matrix = new char[rows][cols];
-    for (int i = 0, idx = 0; i < rows; i++)
-        for (int j = 0; j < cols; j++)
-            matrix[i][j] = array[idx++];
+    for (int r = 0, idx = 0; r < rows; r++)
+        for (int c = 0; c < cols; c++)
+            matrix[r][c] = array[idx++];
     return matrix;
 }
 ```