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From: whusky <31883473+GitHubQAQ@users.noreply.github.com>
Date: Tue, 17 May 2022 21:22:17 +0800
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1. 优化代码高亮格式
2. 对于C++的第一种解法添加预判断代码
---
 problems/0063.不同路径II.md | 6 ++++--
 1 file changed, 4 insertions(+), 2 deletions(-)

diff --git a/problems/0063.不同路径II.md b/problems/0063.不同路径II.md
index a40cceda..0e51e0fe 100644
--- a/problems/0063.不同路径II.md
+++ b/problems/0063.不同路径II.md
@@ -66,7 +66,7 @@ dp[i][j] ：表示从（0 ，0）出发，到(i, j) 有dp[i][j]条不同的路
 
 所以代码为：
 
-```
+```cpp
 if (obstacleGrid[i][j] == 0) { // 当(i, j)没有障碍的时候，再推导dp[i][j]
     dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
 }
@@ -76,7 +76,7 @@ if (obstacleGrid[i][j] == 0) { // 当(i, j)没有障碍的时候，再推导dp[i
 
 在[62.不同路径](https://programmercarl.com/0062.不同路径.html)不同路径中我们给出如下的初始化：
 
-```
+```cpp
 vector<vector<int>> dp(m, vector<int>(n, 0)); // 初始值为0
 for (int i = 0; i < m; i++) dp[i][0] = 1;
 for (int j = 0; j < n; j++) dp[0][j] = 1;
@@ -138,6 +138,8 @@ public:
     int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
         int m = obstacleGrid.size();
         int n = obstacleGrid[0].size();
+	if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 1) //如果在起点或终点出现了障碍，直接返回0
+            return 0;
         vector<vector<int>> dp(m, vector<int>(n, 0));
         for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) dp[i][0] = 1;
         for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) dp[0][j] = 1;