0062. Unique Paths

Author: coderbean

markdown/0062. Unique Paths.md

@@ -0,0 +1,66 @@
+### [62\. Unique Paths](https://leetcode.com/problems/unique-paths/)
+
+Difficulty: **Medium**
+
+
+A robot is located at the top-left corner of a _m_ x _n_ grid (marked 'Start' in the diagram below).
+
+The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
+
+How many possible unique paths are there?
+
+![](https://assets.leetcode.com/uploads/2018/10/22/robot_maze.png)  
+<small style="display: inline;">Above is a 7 x 3 grid. How many possible unique paths are there?</small>
+
+**Note:** _m_ and _n_ will be at most 100.
+
+**Example 1:**
+
+```
+Input: m = 3, n = 2
+Output: 3
+Explanation:
+From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
+1\. Right -> Right -> Down
+2\. Right -> Down -> Right
+3\. Down -> Right -> Right
+```
+
+**Example 2:**
+
+```
+Input: m = 7, n = 3
+Output: 28
+```
+
+
+#### Solution
+
+Language: **Java**
+
+```java
+class Solution {
+    public int uniquePaths(int m, int n) {
+        int[][] ways = new int[m + 1][n + 1]; // 所有的格子都能到达，因此初始化为0，表示没有计算过。该数组用于保存到达该坐标的路径个数，防止重复计算
+        return uniquePaths(ways, m, n);
+    }
+​
+    private int uniquePaths(int[][] ways, int m, int n) {
+        if (ways[m][n] > 0) {
+            return ways[m][n];
+        }
+        if (m * n == 0) {
+            ways[m][n] = 0;
+            return 0;
+        }
+        if (m == 1 || n == 1) {
+            ways[m][n] = 1;
+            return 1;
+        }
+        ways[m][n] = uniquePaths(ways, m, n - 1) + uniquePaths(ways, m - 1, n);
+        return ways[m][n];
+    }
+}
+```
+这里第一次提交超时了，由于没考虑重复计算的问题，造成如果棋盘很大的情况下，会导致重复计算问题明显。
+![](http://ww4.sinaimg.cn/large/006tNc79ly1g51045ecr4j31b40q643d.jpg)

src/main/java/leetcode/_61_/Main.java

@@ -6,7 +6,6 @@
 public class Main {
     public static void main(String[] args) {
         Solution solution = new Solution();
-        System.out.println( solution.isValid(""));
     }
 }
 

src/main/java/leetcode/_62_/Main.java

@@ -0,0 +1,13 @@
+package leetcode._62_;
+
+/**
+ * Created by zhangbo54 on 2019-03-04.
+ */
+public class Main {
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        System.out.println(solution.uniquePaths(7, 3));
+        System.out.println(solution.uniquePaths(51, 9));
+    }
+}
+

src/main/java/leetcode/_62_/Solution.java

@@ -0,0 +1,24 @@
+package leetcode._62_;
+
+class Solution {
+    public int uniquePaths(int m, int n) {
+        int[][] ways = new int[m + 1][n + 1]; // 所有的格子都能到达，因此初始化为0，表示没有计算过。该数组用于保存到达该坐标的路径个数，防止重复计算
+        return uniquePaths(ways, m, n);
+    }
+
+    private int uniquePaths(int[][] ways, int m, int n) {
+        if (ways[m][n] > 0) {
+            return ways[m][n];
+        }
+        if (m * n == 0) {
+            ways[m][n] = 0;
+            return 0;
+        }
+        if (m == 1 || n == 1) {
+            ways[m][n] = 1;
+            return 1;
+        }
+        ways[m][n] = uniquePaths(ways, m, n - 1) + uniquePaths(ways, m - 1, n);
+        return ways[m][n];
+    }
+}

src/main/java/leetcode/_62_/solution.md

@@ -0,0 +1,66 @@
+### [62\. Unique Paths](https://leetcode.com/problems/unique-paths/)
+
+Difficulty: **Medium**
+
+
+A robot is located at the top-left corner of a _m_ x _n_ grid (marked 'Start' in the diagram below).
+
+The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
+
+How many possible unique paths are there?
+
+![](https://assets.leetcode.com/uploads/2018/10/22/robot_maze.png)  
+<small style="display: inline;">Above is a 7 x 3 grid. How many possible unique paths are there?</small>
+
+**Note:** _m_ and _n_ will be at most 100.
+
+**Example 1:**
+
+```
+Input: m = 3, n = 2
+Output: 3
+Explanation:
+From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
+1\. Right -> Right -> Down
+2\. Right -> Down -> Right
+3\. Down -> Right -> Right
+```
+
+**Example 2:**
+
+```
+Input: m = 7, n = 3
+Output: 28
+```
+
+
+#### Solution
+
+Language: **Java**
+
+```java
+class Solution {
+    public int uniquePaths(int m, int n) {
+        int[][] ways = new int[m + 1][n + 1]; // 所有的格子都能到达，因此初始化为0，表示没有计算过。该数组用于保存到达该坐标的路径个数，防止重复计算
+        return uniquePaths(ways, m, n);
+    }
+​
+    private int uniquePaths(int[][] ways, int m, int n) {
+        if (ways[m][n] > 0) {
+            return ways[m][n];
+        }
+        if (m * n == 0) {
+            ways[m][n] = 0;
+            return 0;
+        }
+        if (m == 1 || n == 1) {
+            ways[m][n] = 1;
+            return 1;
+        }
+        ways[m][n] = uniquePaths(ways, m, n - 1) + uniquePaths(ways, m - 1, n);
+        return ways[m][n];
+    }
+}
+```
+这里第一次提交超时了，由于没考虑重复计算的问题，造成如果棋盘很大的情况下，会导致重复计算问题明显。
+![](http://ww4.sinaimg.cn/large/006tNc79ly1g51045ecr4j31b40q643d.jpg)