leetcode 62 递归的倒推和数组的正推 leetcode63 障碍物补充

写这个的目的是为了方便以后回顾，这道题的总结在于迭代时间代价比较大 

题目要求：

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.

这道题的一个问题是它的下坐标相当于从（1，1）开始，也踩了一次坑

看到这道题我的思路是迭代，于是很想当然的就掉坑了，迭代的代码

public int uniquePaths(int m, int n) {
		if(n<1||m<1){
			return 0;
		}
		if(n==1||m==1){
            return 1;
        }
		
		int s1=uniquePaths(m,n-1);
		int s2=uniquePaths(m-1,n);
		return s1+s2;
	}

提交上去时间超时

当时学数据结构的时候老师就讲过这两者的区别，想朱老师呀！

正推的代码

public int uniquePaths(int m, int n) {
		int[][] temp=new int[m+1][n+1];
		for(int i=1;i<=m;i++){			
			temp[i][1]=1;
		}
		for(int j=1;j<=n;j++){
			temp[1][j]=1;
		}
		for(int i=2;i<=m;i++){
			for(int j=2;j<=n;j++){
				temp[i][j]=temp[i-1][j]+temp[i][j-1];
			}
		}
		return temp[m][n];
	}

这个错误我前面貌似犯过呀！看来还需要多记

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).

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.

Note: m and n will be at most 100.

Example 1:

Input:
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

解决办法：

package test;

public class LC63Try1
{
	public int uniquePathsWithObstacles(int[][] obstacleGrid)
	{
		int m = obstacleGrid[0].length;
		int n = obstacleGrid.length;
		int[] temp = new int[m];
		temp[0] = 1;
		if(obstacleGrid[0][0] == 1){
			temp[0] = 0;
		}
		for (int i = 1; i < m; i++)
		{
			if (obstacleGrid[0][i] == 1)
			{
				temp[i] = 0;
			}
			else
			{
				temp[i] = temp[i - 1];
			}

		}
		for (int j = 1; j < n; j++)
		{
			if (obstacleGrid[j][0] == 1)
			{
				temp[0] = 0;
			}
			for (int i = 1; i < m; i++)
			{
				if (obstacleGrid[j][i] == 1)
				{
					temp[i] = 0;
				}
				else
				{
					temp[i] = temp[i - 1] + temp[i];
				}
			}

		}
		return temp[m - 1];

	}

	public static void main(String[] args)
	{
		LC63Try1 Object1 = new LC63Try1();
		int[][] obstacleGrid = { { 0, 0, 0 }, { 0, 1, 0 }, { 0, 0, 0 } };
		System.out.println(Object1.uniquePathsWithObstacles(obstacleGrid));

	}

}

哈哈哈