LeetCode Unique Paths

原题链接在这里：https://leetcode.com/problems/unique-paths/

题目：

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?

Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

题解：

DP问题.需保存历史数据为走到当前格子的不同路径数，用二维数组dp保存。

更新当前点dp[i][j]为上一行同列dp[i-1][j]的值 + 本行上一列dp[i][j-1]的值，因为走到上一行同列的值想走到当前格都是往下走一步，左边同理。

初始化是第一行和第一列都是1.

答案是dp[m-1][n-1].

Time Complexity: O(m*n). Space: O(m*n).

AC Java:

 1 public class Solution {
 2     public int uniquePaths(int m, int n) {
 3         if(m == 0 || n == 0){
 4             return 0;
 5         }
 6         int [][] dp = new int[m][n];
 7         for(int i = 0; i<m; i++){
 8             dp[i][0] = 1;
 9         }
10         for(int j = 0; j<n; j++){
11             dp[0][j] = 1;
12         }
13         for(int i = 1; i<m; i++){
14             for(int j = 1; j<n; j++){
15                 dp[i][j] = dp[i-1][j] + dp[i][j-1];
16             }
17         }
18         return dp[m-1][n-1];
19     }
20 }

存储历史信息可以用一维数组完成从而节省空间。生成一个长度为n的数组dp, 每次更新dp[j] += dp[j-1], dp[j-1]就是同行前一列的历史结果，dp[j]为更新前是同列上一行的结果，所以dp[j] += dp[j-1]就是更新后的结果。

Note:外层loop i 从0开始，dp[0] = 1, 相当于初始了第一列，所以i=0开始要初始第一行

Time Complexity: O(m*n). Space: O(n).

 1 class Solution {
 2     public int uniquePaths(int m, int n) {
 3         if(m == 0 || n == 0){
 4             return 0;
 5         }
 6
 7         int [] dp = new int[n];
 8         for(int j = 0; j<n; j++){
 9             dp[j] = 1;
10         }
11
12         for(int i = 1; i<m; i++){
13             for(int j = 1; j<n; j++){
14                 dp[j] = dp[j] + dp[j-1];
15             }
16         }
17
18         return dp[n-1];
19     }
20 }

有进阶版题目Unique Paths II.

转载于:https://www.cnblogs.com/Dylan-Java-NYC/p/4824960.html

LeetCode Unique Paths相关推荐

LeetCode | Unique Paths I,II
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The ...
[leetcode]Unique Paths II
>这道题目和上一题基本是一样的.只需要检测现在正在处理的那个单位是不是不可通过的,如果是就直接让这个单元格=0,也就是没有路径可以到右下角就好了. 代码如下: public class Solu ...
[Lintcode]115. Unique Paths II/[Leetcode]63. Unique Paths II
115. Unique Paths II/63. Unique Paths II 本题难度: Easy/Medium Topic: Dynamic Programming Description Fo ...
leetcode 62, 63, 980. Unique Paths I, II, III | 62, 63, 980. 不同路径 I, II, III（暴力递归-＞傻缓存-＞动态规划）
62. Unique Paths https://leetcode.com/problems/unique-paths/ 注意本题只能向右 / 向上走. DP 问题,经典又熟悉. 暴力递归->傻 ...
【动态规划】LeetCode 63. Unique Paths II
LeetCode 63. Unique Paths II Solution1:我的答案在哪里做过这题? class Solution { public:int uniquePathsWithObst ...
【动态规划】LeetCode 62. Unique Paths
LeetCode 62. Unique Paths Solution1:我的未能AC的答案递归超时了!!! class Solution { public:int uniquePaths(int m ...
LeetCode 1.Minimum Path Sum 2.Unique Paths I and II
大家好,我是刘天昊,快到端午节了,今天说两道动态规划的题目(话说动规真的挺难的) 当然这三题是一样的解体思路先看Unique Paths A robot is located at the top-l ...
Leetcode 62. Unique Paths
题目: A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). ...
Unique Paths II leetcode java
题目: Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. H ...

LeetCode Unique Paths

LeetCode Unique Paths相关推荐

最新文章

热门文章