From 6a4eba94df4e24e3f3284d55399290c23e13ee57 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <tangshusen@pku.edu.cn>
Date: Mon, 18 Feb 2019 23:28:44 +0800
Subject: [PATCH] Update 63. Unique Paths II.md

---
 solutions/63. Unique Paths II.md | 21 +++++++++++----------
 1 file changed, 11 insertions(+), 10 deletions(-)

diff --git a/solutions/63. Unique Paths II.md b/solutions/63. Unique Paths II.md
index e293f53..bb9f53a 100644
--- a/solutions/63. Unique Paths II.md	
+++ b/solutions/63. Unique Paths II.md	
@@ -2,7 +2,9 @@
 # 思路
 这题其实和[62. Unique Paths](https://leetcode.com/problems/unique-paths/)几乎一样的，唯一区别就是矩形网格里存在一些障碍物，我们对每个格子判断一下就行了。
 就是一个简单的动归，可参考[62题题解](https://github.com/ShusenTang/LeetCode/blob/master/solutions/62.%20Unique%20Paths.md)。     
-时间空间复杂度都是O(mn)
+时间空间复杂度都是O(mn)     
+另外，类似62题，可将空间复杂度优化到O(m)，这里不再赘述，可参考[讨论区](https://leetcode.com/problems/unique-paths-ii/discuss/23252/4ms-O(n)-DP-Solution-in-C%2B%2B-with-Explanations)         
+注意: dp数组开成int型时有的case过不了（之前是没有这些case的，直接int就能过），所以开成long long型的
 
 # C++
 ``` C++
@@ -10,14 +12,13 @@ class Solution {
 public:
     int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
         int m = obstacleGrid.size(), n = obstacleGrid[0].size();
-        vector<vector<int>>dp(m + 1, vector<int>(n + 1, 0));
-        dp[1][1] = 1;
-        for(int i = 1; i <= m; i++)
-            for(int j = 1; j <= n; j++){
-                if(obstacleGrid[i - 1][j - 1]) dp[i][j] = 0; // 此处为障碍物
-                else dp[i][j] += (dp[i - 1][j] + dp[i][j - 1]);
-            }
-        return dp[m][n];
-    }
+        vector<vector<long long> > dp(m + 1, vector<long long> (n + 1, 0));
+        dp[0][1] = 1;
+        for (int i = 1; i <= m; i++)
+            for (int j = 1; j <= n; j++)
+                if (!obstacleGrid[i - 1][j - 1])
+                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
+        return int(dp[m][n]);
+    } 
 };
 ```