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  • leetcode 63 Unique Paths II ---java

    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.

    Note: m and n will be at most 100.

    这道题就是上一道题的延伸,相当于上一道题就是给定了一个m*n的矩阵,而且全是0,这里位置1表示的不能走的地方。所以在上一个版本上修正。

    区别就是在于如果遇到1的时候的处理方式不同。

    public class Solution {
        public int uniquePathsWithObstacles(int[][] obstacleGrid) {
            int m = obstacleGrid.length;
            int n = obstacleGrid[0].length;
            if( m == 0 || n == 0)
                return 0;
            if( m == 1){
                for( int i = 0;i<n;i++){
                    if( obstacleGrid[0][i] == 1 )
                        return 0;
                }
                return 1;
            }
            if( n == 1){
                for( int i = 0;i<m;i++){
                    if( obstacleGrid[i][0] == 1)
                        return 0;
                }
                return 1;
            }
            int[] dp = new int[n];
            for( int i = 0;i<n;i++){
                if( obstacleGrid[0][i] == 1 )
                    while(i<n){
                        dp[i] = 0;
                        i++;
                }
                else
                    dp[i] = 1;
            }
            for( int i = 1;i < m;i++){
                if( obstacleGrid[i][0] == 1){
                    int j = 0;
                    while(j<n){
                            if( obstacleGrid[i][j] == 1)
                                dp[j] = 0;
                            else
                                dp[j] = dp[j];
                            j++;
                        }
                }
                for( int j = 1;j<n;j++){
                    if( obstacleGrid[i][j] == 1){
                        dp[j] = 0;
                    }else
                        dp[j] += dp[j-1];
                }
            }
            return dp[n-1];
            
            
            
            
            
            
        }
    }
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  • 原文地址:https://www.cnblogs.com/xiaoba1203/p/5947748.html
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