From e537f56200443e95feb7a59c9d80d008bd103b1b Mon Sep 17 00:00:00 2001
From: lichuanming <1465802096@qq.com>
Date: Thu, 27 May 2021 10:39:06 +0800
Subject: [PATCH] =?UTF-8?q?java=E7=89=88=E6=9C=AC=20lC62.=E4=B8=8D?=
 =?UTF-8?q?=E5=90=8C=E8=B7=AF=E5=BE=84?=
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---
 problems/0062.不同路径.md | 30 ++++++++++++++++++++++++++++++
 1 file changed, 30 insertions(+)

diff --git a/problems/0062.不同路径.md b/problems/0062.不同路径.md
index c4673a23..5fe0c850 100644
--- a/problems/0062.不同路径.md
+++ b/problems/0062.不同路径.md
@@ -243,7 +243,37 @@ public:
 
 
 Java：
+```java
+  /**
+     * 1. 确定dp数组下表含义 dp[i][j] 到每一个坐标可能的路径种类
+     * 2. 递推公式 dp[i][j] = dp[i-1][j] dp[i][j-1]
+     * 3. 初始化 dp[i][0]=1 dp[0][i]=1 初始化横竖就可
+     * 4. 遍历顺序 一行一行遍历
+     * 5. 推导结果 。。。。。。。。
+     *
+     * @param m
+     * @param n
+     * @return
+     */
+    public static int uniquePaths(int m, int n) {
+        int[][] dp = new int[m][n];
+        //初始化
+        for (int i = 0; i < m; i++) {
+            dp[i][0] = 1;
+        }
+        for (int i = 0; i < n; i++) {
+            dp[0][i] = 1;
+        }
 
+        for (int i = 1; i < m; i++) {
+            for (int j = 1; j < n; j++) {
+                dp[i][j] = dp[i-1][j]+dp[i][j-1];
+            }
+        }
+        return dp[m-1][n-1];
+    }
+
+```
 
 Python：