LeetCode之动态规划

62. Unique Paths

QuestionEditorial Solution

Total Accepted: 86710 Total Submissions: 239084 Difficulty: Medium

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?

使用动态规划来求解本题，在每个节点处，只有两个选择，向下或者向右走。本节点可由上一个节点向下或者向右走过来，两种方法：

1）使用一个二位数组count[i][j]来表示从起始点到达坐标(i,j)的，有两种，（i-1,j）-->(i,j)及(i,j-1)-->（i,j），即count[i][j] = count[i-1][j]+count[i][j-1]，此时时间复杂度为O（mn），空间复杂度为O(mn)

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

2）使用一个数组count[i]来表示从起始点到达某一行某一列的路径数量，此时有，count[i] = count[i-1]+count[i]，此时，时间复杂度为O（mn），空间复杂度为O(n)：

//因为只需求得最终结果，而不需要知道到达每一个坐标的路径数量public class Solution {
    public int uniquePaths(int m, int n) {
        int[] count = new int[n];
        count[0]=1;
        for(int i=0;i<m;i++){
            for(int j=1;j<n;j++){
                count[i] = count[i-1]+count[i];
            }
        }
        return count[n-1];
    }
}

63. Unique Paths II

QuestionEditorial Solution

Total Accepted: 65258 Total Submissions: 222092 Difficulty: Medium

Follow up for "Unique Paths":

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

The total number of unique paths is 2.