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[LeetCode][Medium] 62. Unique Paths

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

題目

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?

Example 1:

alt

Input: m = 3, n = 7
Output: 28

Example 2:

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 -> Down -> Down
2. Down -> Down -> Right
3. Down -> Right -> Down

Example 3:

Input: m = 7, n = 3
Output: 28

Example 4:

Input: m = 3, n = 3
Output: 6

思路

dp[i][j] 代表到達該點的路徑數量。而該點可以從左邊或上面到達 也就是 dp[i][j] = dp[i - 1][j] + dp[i][j - 1] 。

Solution

/**
 * @param {number} m
 * @param {number} n
 * @return {number}
 */
var uniquePaths = function(m, n) {
    var dp = new Array(m)
    
    if (!m || !n) return 0
    
    for (var i = 0; i < m; i++) {
        dp[i] = new Array(n)
        for (var j = 0; j < n; j++) {
            if (j > 0 && i > 0) dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
            else if (j > 0 && i === 0) dp[i][j] = dp[i][j - 1]
            else if (j === 0 && i > 0) dp[i][j] = dp[i - 1][j]
            else dp[i][j] = 1
        }
    }
    return dp[m - 1][n - 1]
};

Complexity

  • Time complexity : O(m * n)
  • Space complexity : O(m * n)

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