From 5fc3a4e7458fc415807b64a450fb24744c671ff5 Mon Sep 17 00:00:00 2001
From: ShusenTang <tangshusen@pku.edu.cn>
Date: Mon, 10 Feb 2020 20:40:32 +0800
Subject: [PATCH] add 64

---
 README.md                         |  1 +
 solutions/64. Minimum Path Sum.md | 57 +++++++++++++++++++++++++++++++
 2 files changed, 58 insertions(+)
 create mode 100644 solutions/64. Minimum Path Sum.md

diff --git a/README.md b/README.md
index f30b21d..1a50104 100644
--- a/README.md
+++ b/README.md
@@ -57,6 +57,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 61 |[Rotate List](https://leetcode.com/problems/rotate-list/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/61.%20Rotate%20List.md)|Medium|  |
 | 62 |[Unique Paths](https://leetcode.com/problems/unique-paths/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/62.%20Unique%20Paths.md)|Medium|  |
 | 63 |[Unique Paths II](https://leetcode.com/problems/unique-paths-ii/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/63.%20Unique%20Paths%20II.md)|Medium|  |
+| 64 |[Minimum Path Sum](https://leetcode.com/problems/minimum-path-sum/)|[C++](solutions/64.%20Minimum%20Path%20Sum.md)|Medium|  |
 | 65 |[Valid Number](https://leetcode.com/problems/valid-number/)|[C++](solutions/65.%20Valid%20Number.md)|Hard|  |
 | 66 |[Plus One](https://leetcode.com/problems/plus-one)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/66.%20Plus%20One.md)|Easy|  |
 | 67 |[Add Binary](https://leetcode.com/problems/add-binary)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/67.%20Add%20Binary.md)|Easy|  |
diff --git a/solutions/64. Minimum Path Sum.md b/solutions/64. Minimum Path Sum.md
new file mode 100644
index 0000000..c9df0d2
--- /dev/null
+++ b/solutions/64. Minimum Path Sum.md	
@@ -0,0 +1,57 @@
+# [64. Minimum Path Sum](https://leetcode.com/problems/minimum-path-sum/)
+
+# 思路
+
+给定一个`m x n`的矩阵，从左上角走到右下角，求最小路径和。就是一个简单的动态规划：
+```
+dp[i][j] 代表从 grid[0][0] 走到 grid[i][j] 的最小路径和。
+```
+因为只能往下或者往右走，所以状态更新方程为
+```
+dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1]);
+```
+
+可以利用滚动数组的方式将空间复杂度优化到O(n)或者O(m)。
+
+
+值得一提的是如果给定的矩阵grid可写的话我们可以就将其作为dp数组，这样就不用额外的空间了。
+
+# C++
+
+``` C++
+class Solution {
+public:
+    int minPathSum(vector<vector<int>>& grid) {
+        int m = grid.size(), n = grid[0].size();
+        
+        vector<vector<int>>dp(m, vector<int>(n, INT_MAX));
+        dp[0][0] = grid[0][0];
+        for(int i = 1; i < m; i++) dp[i][0] = dp[i-1][0] + grid[i][0];
+        for(int j = 1; j < n; j++) dp[0][j] = dp[0][j-1] + grid[0][j];
+        
+        for(int i = 1; i < m; i++)
+            for(int j = 1; j < n; j++)
+                dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1]);
+        
+        return dp[m-1][n-1];
+    }
+};
+```
+
+## 滚动数组空间优化
+``` C++
+class Solution {
+public:
+    int minPathSum(vector<vector<int>>& grid) {
+        int m = grid.size(), n = grid[0].size();
+        
+        vector<int>dp(n, INT_MAX);
+        dp[0] = 0;
+        for(int i = 0; i < m; i++)
+            for(int j = 0; j < n; j++)
+                dp[j] = grid[i][j] + min(j > 0 ? dp[j-1] : INT_MAX, dp[j]);
+        
+        return dp.back();
+    }
+};
+```
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