[LeetCode] Unique Paths II

最新推荐文章于 2019-05-17 12:08:12 发布

原创最新推荐文章于 2019-05-17 12:08:12 发布 · 153 阅读

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本文介绍了一种解决网格中存在障碍物时寻找从起点到终点的唯一路径数量的方法。采用动态规划算法，在原数组上进行修改以节省空间，实现O(1)的空间复杂度。

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.

动态规划：o(1)空间复杂度在原数组上修改

public class Solution {
	//o(1)空间复杂度  在原数组上修改
	public int uniquePathsWithObstacles(int[][] obstacleGrid) {
		if(obstacleGrid.length==0||obstacleGrid[0].length==0) return 0;
		for(int i=0;i<obstacleGrid.length;i++){
			for(int j=0;j<obstacleGrid[0].length;j++){
				if(obstacleGrid[i][j]==1) obstacleGrid[i][j]=-1;
			}
		}
		for(int i=0;i<obstacleGrid[0].length;i++){
			if(obstacleGrid[0][i]==-1) obstacleGrid[0][i]=0;
			else if(i==0) obstacleGrid[0][i]=1;
			else obstacleGrid[0][i]=obstacleGrid[0][i-1];
		}
		for(int i=1;i<obstacleGrid.length;i++){
			if(obstacleGrid[i][0]==-1) obstacleGrid[i][0]=0;
			else obstacleGrid[i][0]=obstacleGrid[i-1][0];
		}
		for(int i=1;i<obstacleGrid.length;i++){
			for(int j=1;j<obstacleGrid[0].length;j++){
				if(obstacleGrid[i][j]==-1) obstacleGrid[i][j]=0;
				else  obstacleGrid[i][j]=obstacleGrid[i-1][j]+obstacleGrid[i][j-1];
			}
		}
		return obstacleGrid[obstacleGrid.length-1][obstacleGrid[0].length-1];
	}
}