[LintCode/LeetCode] Unique Paths

Gu_Yan 發(fā)布于2019-08-14 15:27 / 2774人閱讀

摘要：簡單的動規(guī)題目，建立數(shù)組。坐標為矩陣的坐標，值為從左上角到這一格的走法總數(shù)。賦初值，最上一行和最左列的所有格子的走法都只有一種，其余格子的走法等于其左邊格子走法與上方格子走法之和。最后，返回即可。

Problem

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?

Note

簡單的動規(guī)題目，建立dp數(shù)組。dp[i][j]坐標為矩陣的坐標，值為從左上角到這一格的走法總數(shù)。賦初值，最上一行和最左列的所有格子的走法都只有一種，其余格子的走法等于其左邊格子走法與上方格子走法之和。最后，返回dp[m-1][n-1]即可。

Solution

二維DP

public class Solution {
    public int uniquePaths(int m, int n) {
        int[][] dp = new int[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0 || j == 0) dp[i][j] = 1;
                else dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }
        return dp[m-1][n-1];
    }
}

一維DP

public class Solution {
    public int uniquePaths(int m, int n) {
        int[] dp = new int[n];
        Arrays.fill(dp, 1);
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[j] += dp[j-1];
            }
        }
        return dp[n-1];
    }
}

云服務器 GPU云服務器 Paths unique Mysql_unique array_unique

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