From 4f4a54fa7928a11d33a498310426793daeabab77 Mon Sep 17 00:00:00 2001
From: ArthurP <gpan20020205@gmail.com>
Date: Sun, 16 Jan 2022 09:50:17 +0000
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200063.=E4=B8=8D=E5=90=8C?=
 =?UTF-8?q?=E8=B7=AF=E5=BE=84II.md=20C=E8=AF=AD=E8=A8=80=E7=89=88=E6=9C=AC?=
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---
 problems/0063.不同路径II.md | 54 +++++++++++++++++++++++++++++++++
 1 file changed, 54 insertions(+)

diff --git a/problems/0063.不同路径II.md b/problems/0063.不同路径II.md
index 490b6b5c..6f405d6a 100644
--- a/problems/0063.不同路径II.md
+++ b/problems/0063.不同路径II.md
@@ -333,6 +333,60 @@ var uniquePathsWithObstacles = function(obstacleGrid) {
 };
 ```
 
+C
+```c
+//初始化dp数组
+int **initDP(int m, int n, int** obstacleGrid) {
+    int **dp = (int**)malloc(sizeof(int*) * m);
+    int i, j;
+    //初始化每一行数组
+    for(i = 0; i < m; ++i) {
+        dp[i] = (int*)malloc(sizeof(int) * n);
+    }
+
+    //先将第一行第一列设为0
+    for(i = 0; i < m; ++i) {
+        dp[i][0] = 0;
+    }
+    for(j = 0; j < n; ++j) {
+        dp[0][j] = 0;
+    }
+
+    //若碰到障碍，之后的都走不了。退出循环
+    for(i = 0; i < m; ++i) {
+        if(obstacleGrid[i][0]) {
+            break;
+        }
+        dp[i][0] = 1;
+    }
+    for(j = 0; j < n; ++j) {
+        if(obstacleGrid[0][j])
+            break;
+        dp[0][j] = 1;
+    }
+    return dp;
+}
+
+int uniquePathsWithObstacles(int** obstacleGrid, int obstacleGridSize, int* obstacleGridColSize){
+    int m = obstacleGridSize, n = *obstacleGridColSize;
+    //初始化dp数组
+    int **dp = initDP(m, n, obstacleGrid);
+
+    int i, j;
+    for(i = 1; i < m; ++i) {
+        for(j = 1; j < n; ++j) {
+            //若当前i,j位置有障碍
+            if(obstacleGrid[i][j])
+                //路线不同
+                dp[i][j] = 0;
+            else
+                dp[i][j] = dp[i-1][j] + dp[i][j-1];
+        }
+    }
+    //返回最后终点的路径个数
+    return dp[m-1][n-1];
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>