LeetCode 62. Unique Paths Java
阿新 • • 發佈:2017-05-28
fin can 一個 其中 leet logs 網格 marked int
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?
Above is a 3 x 7 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
題意:給出一個m*n的網格,有一個機器人,每一步只能向右或者向下走,求能從最左上角的格子到最右下角的格子有多少種走法。這道題是典型的動態規劃,將網格看成一個二維數組A[m][n],那麽我們要求的就是從A[0][0]到A[m-1][n-1]的路徑,對於其中任意一點A[i][j](i<m,j<n)來說,到該點的方法有從A[i-1][j]向下走一步或者A[i][j-1]向右走一步,所以d(A[i][j])=d(A[i-1][j])+d(A[i][j-1])。對每個點遍歷求解即可。這裏需要註意對第一行和第一列上的所有點都只有一種走法。
代碼:
public class Solution { public int uniquePaths(int m, int n) { if(m==0||n==0) return 0; if(m==1||n==1) //只有一行或者只有一列 只有一種走法 return 1; int[][] A=new int[m][n]; //用戶記錄起點到當前點走法 for(int i=0;i<m;i++){ for(int j=0;j<n;j++){ if(i==0||j==0) A[i][j]=1; else A[i][j]=A[i-1][j]+A[i][j-1]; } } return A[m-1][n-1]; } }
LeetCode 62. Unique Paths Java