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Difficulty: 中等
Related Topics: 数学, 动态规划, 组合数学
一个机器人位于一个 m x n网格的左上角 (起始点在下图中标记为 “Start” )。
m x n
机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角(在下图中标记为 “Finish” )。
问总共有多少条不同的路径?
示例 1:
输入:m = 3, n = 7 输出:28
示例 2:
输入:m = 3, n = 2 输出:3 解释: 从左上角开始,总共有 3 条路径可以到达右下角。 1\. 向右 -> 向下 -> 向下 2\. 向下 -> 向下 -> 向右 3\. 向下 -> 向右 -> 向下
示例 3:
输入:m = 7, n = 3 输出:28
示例 4:
输入:m = 3, n = 3 输出:6
提示:
1 <= m, n <= 100
Language: JavaScript
/** * @param {number} m * @param {number} n * @return {number} * 思路: * 每一格的路径由其上一格和左一格决定。 * 方法:动态规划 * 动态方程:dp[i][j] = dp[i-1][j] + dp[i][j-1] * 注意:对于第一行 dp[0][j],或者第一列 dp[i][0],由于都是在边界,所以只能为1 * 建立m*n的矩阵,注意第0行和第0列元素均为1 */ var uniquePaths = function(m, n) { const dp = new Array(m).fill(0).map(() => new Array(n).fill(0)) for (let i = 0; i < m; i++) { dp[i][0] = 1 } for (let j = 0; j < n; j++) { dp[0][j] = 1 } for (let i = 1; i < m; i++) { for (let j = 1; j < n; j++) { dp[i][j] = dp[i-1][j] + dp[i][j-1] } } return dp[m-1][n-1] }
The text was updated successfully, but these errors were encountered:
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62. 不同路径
Description
Difficulty: 中等
Related Topics: 数学, 动态规划, 组合数学
一个机器人位于一个
m x n
网格的左上角 (起始点在下图中标记为 “Start” )。机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角(在下图中标记为 “Finish” )。
问总共有多少条不同的路径?
示例 1:
示例 2:
示例 3:
示例 4:
提示:
1 <= m, n <= 100
Solution
Language: JavaScript
The text was updated successfully, but these errors were encountered: