UniquePaths.java

Author: 李文超

README.md

@@ -84,6 +84,7 @@ This is leetcode solution in Java with eplaination and generalization.
 | # | Title | Solution | Difficulty | Source Code |
 |---| ----- | -------- | ---------- | ----------- |
 | 5 | [Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/) | dp | Medium | [LongestPalindromicSubstring.java](https://github.com/venciallee/leetcode-in-Java/blob/master/algorithm/app/src/main/java/com/bytecode/leetcode/dp/LongestPalindromicSubstring.java) |
+| 62 | [Unique Paths](https://leetcode.com/problems/unique-paths/) | dp | Medium | [UniquePaths.java](https://github.com/venciallee/leetcode-in-Java/blob/master/algorithm/app/src/main/java/com/bytecode/leetcode/dp/UniquePaths.java) |
 
 ## Design
 

algorithm/app/src/main/java/com/bytecode/leetcode/dp/UniquePaths.java

@@ -0,0 +1,66 @@
+package com.bytecode.leetcode.dp;
+
+/**
+ * 62. Unique Paths
+ * <p>
+ * A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
+ * <p>
+ * 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).
+ * <p>
+ * How many possible unique paths are there?
+ * <p>
+ * <p>
+ * Above is a 7 x 3 grid. How many possible unique paths are there?
+ * <p>
+ * Note: m and n will be at most 100.
+ * <p>
+ * Example 1:
+ * <p>
+ * 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:
+ * <p>
+ * Input: m = 7, n = 3
+ * Output: 28
+ * <p>
+ * <a href="https://leetcode.com/problems/unique-paths/">62. Unique Paths</a>
+ * <p>
+ * <a href="https://leetcode-cn.com/problems/unique-paths/">62. 不同路径</a>
+ * <p>
+ * Created by vencial on 2019-09-10.
+ */
+public class UniquePaths {
+
+    /**
+     * 1. 设dp为每格可能路径的总数, 则第一行和第一列均为1
+     * 2. dp[i][j]则为dp[i - 1][j] 加上dp[i][j - 1]的总和
+     * <p>
+     * 1. let dp is the sum of possible unique paths, as we know dp of the first row and column is 1
+     * 2. dp[i][j] is the sum of left dp and the top dp, that is dp[i - 1][j] + dp[i][j - 1]
+     *
+     * @param m m
+     * @param n n
+     * @return int
+     */
+    public int uniquePaths(int m, int n) {
+        int dp[][] = new int[m][n];
+        dp[0][0] = 0;
+        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];
+    }
+}