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?
因为只能向下或者向右走,所依某一点的值就是上面方格的值 和左边的值 相加
dynamic programming.
public class Solution { public int uniquePaths(int m, int n) { int[][] grid=new int[100][100]; for(int i=0;i<m;i++){ grid[i][0]=1; } for(int j=0;j<n;j++){ grid[0][j]=1; } for(int i=1;i<m;i++){ for(int j=1;j<n;j++){ grid[i][j]=grid[i-1][j]+grid[i][j-1]; } } return grid[m-1][n-1]; } }
Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.
Note: You can only move either down or right at any point in time.
public class Solution { public int minPathSum(int[][] grid) { if(grid==null||grid.length==0||grid[0].length==0){ return 0; } int m=grid.length; int n=grid[0].length; for(int i=1;i<n;i++){ grid[0][i]=grid[0][i-1]+grid[0][i]; } for(int j=1;j<m;j++){ grid[j][0]=grid[j-1][0]+grid[j][0]; } for(int i=1;i<m;i++){ for(int j=1;j<n;j++){ if(grid[i-1][j]<=grid[i][j-1]){ grid[i][j]=grid[i-1][j]+grid[i][j]; } else{ grid[i][j]=grid[i][j-1]+grid[i][j]; } } } return grid[grid.length-1][grid[0].length-1]; } }