From 0c184a5ee6e3f52d32e23a0100a5215512f8517b Mon Sep 17 00:00:00 2001
From: Guanzhong Pan <gpan20020205@gmail.com>
Date: Thu, 13 Jan 2022 14:04:08 +0000
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200062.=E4=B8=8D=E5=90=8C?=
 =?UTF-8?q?=E8=B7=AF=E5=BE=84.md=20C=E8=AF=AD=E8=A8=80=E7=89=88=E6=9C=AC?=
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---
 problems/0062.不同路径.md | 36 +++++++++++++++++++++++++++++++++++
 1 file changed, 36 insertions(+)

diff --git a/problems/0062.不同路径.md b/problems/0062.不同路径.md
index e209abb1..efa85a03 100644
--- a/problems/0062.不同路径.md
+++ b/problems/0062.不同路径.md
@@ -347,6 +347,42 @@ var uniquePaths = function(m, n) {
 };
 ```
 
+### C
+```c
+//初始化dp数组
+int **initDP(int m, int n) {
+    //动态开辟dp数组
+    int **dp = (int**)malloc(sizeof(int *) * m);
+    int i, j;
+    for(i = 0; i < m; ++i) {
+        dp[i] = (int *)malloc(sizeof(int) * n);
+    }
+
+    //从0，0到i,0只有一种走法，所以dp[i][0]都是1，同理dp[0][j]也是1
+    for(i = 0; i < m; ++i) 
+        dp[i][0] = 1;
+    for(j = 0; j < n; ++j) 
+        dp[0][j] = 1;
+    return dp;
+}
+
+int uniquePaths(int m, int n){
+    //dp数组，dp[i][j]代表从dp[0][0]到dp[i][j]有几种走法
+    int **dp = initDP(m, n);
+
+    int i, j;
+    //到达dp[i][j]只能从dp[i-1][j]和dp[i][j-1]出发
+    //dp[i][j] = dp[i-1][j] + dp[i][j-1]
+    for(i = 1; i < m; ++i) {
+        for(j = 1; j < n; ++j) {
+            dp[i][j] = dp[i-1][j] + dp[i][j-1];
+        }
+    }
+    int result = dp[m-1][n-1];
+    free(dp);
+    return result;
+}
+```
 
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 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>