zoukankan      html  css  js  c++  java
  • [LeetCode] 63.Unique Paths II

    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

    62. Unique Paths 的拓展, 有以下不同:

    1. 当(i, j)有障碍时dp[i][j] = 0
    2. dp[0][j]和dp[i][0]未必为1.
    dp[0][j] = obstacleGrid[0][j] ? 0 : dp[0][j-1]
    dp[i][0] = obstacleGrid[i][0] ? 0 : dp[i-1][0]
    3. 当obstacleGrid [0][0] = 1时,return 0
    解法:DP, 建立二维数组,dp[i][j]表示在某一位置能到达的不同路径数量,dp[i][j] = dp[i-1][j] + dp[i][j-1] ,如果某一位置grid[i][j]=1说明为障碍物,那么dp[i][j] = 0,还要注意i, j为0的情况。
     
    Java: 
    class Solution {
        public int uniquePathsWithObstacles(int[][] obstacleGrid) { 
            int m = obstacleGrid.length; 
            int n = obstacleGrid[0].length; 
    
            if (m == 0 || n == 0) { 
                return 0;
            }
    
            if (obstacleGrid[0][0] == 1 || obstacleGrid[m-1][n-1] == 1) {
                return 0;
            }
    
            int[][] dp = new int[m][n];
    
            dp[0][0] = 1; 
            for(int i = 1; i < n; i++){ 
                if(obstacleGrid[0][i] == 1) {
                    dp[0][i] = 0; 
                }
                else {
                    dp[0][i] = dp[0][i-1]; 
                }
            } 
    
            for(int i = 1; i < m; i++){ 
                if(obstacleGrid[i][0] == 1) {
                    dp[i][0] = 0; 
                }
                else {
                    dp[i][0] = dp[i-1][0]; 
                }
            } 
    
            for(int i = 1; i < m; i++){ 
                for(int j = 1; j < n; j++){ 
                    if(obstacleGrid[i][j] == 1) {
                        dp[i][j] = 0; 
                    }
                    else {
                        dp[i][j] = dp[i][j-1] + dp[i-1][j]; 
                    }
                } 
            } 
            return dp[m-1][n-1]; 
        }
    }

    Java:

    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int width = obstacleGrid[0].length;
        int[] dp = new int[width];
        dp[0] = 1;
        for (int[] row : obstacleGrid) {
            for (int j = 0; j < width; j++) {
                if (row[j] == 1)
                    dp[j] = 0;
                else if (j > 0)
                    dp[j] += dp[j - 1];
            }
        }
        return dp[width - 1];
    }  

    Python:

    class Solution:
    
        def uniquePathsWithObstacles(self, obstacleGrid):
            mp = obstacleGrid
            for i in range(len(mp)):
                for j in range(len(mp[i])):
                    if i == 0 and j == 0:
                        mp[i][j] = 1 - mp[i][j]
                    elif i == 0:
                        if mp[i][j] == 1:
                            mp[i][j] = 0
                        else:
                            mp[i][j] = mp[i][j - 1]
                    elif j == 0:
                        if mp[i][j] == 1:
                            mp[i][j] = 0
                        else:
                            mp[i][j] = mp[i - 1][j]
                    else:
                        if mp[i][j] == 1:
                            mp[i][j] = 0
                        else:
                            mp[i][j] = mp[i - 1][j] + mp[i][j - 1]
            if mp[-1][-1] > 2147483647: 
                return -1
            else:
                return mp[-1][-1]
    

    Python: wo

    class Solution(object):
        def uniquePathsWithObstacles(self, obstacleGrid):
            """
            :type obstacleGrid: List[List[int]]
            :rtype: int
            """
            m, n = len(obstacleGrid), len(obstacleGrid[0])
            dp = [[0] * n for i in xrange(m)]
            for i in xrange(m):
                for j in xrange(n):
                    if obstacleGrid[i][j] == 1:
                        dp[i][j] == 0
                    else:
                        if i == 0 and j == 0:  # 写错成了 or 
                            dp[i][j] = 1
                        elif i == 0:
                            dp[i][j] = dp[i][j-1]
                        elif j == 0:
                            dp[i][j] = dp[i-1][j]                        
                        else:
                            dp[i][j] = dp[i-1][j] + dp[i][j-1]
                            
            return dp[-1][-1]    

    Python: wo

    class Solution(object):
        def uniquePathsWithObstacles(self, obstacleGrid):
            """
            :type obstacleGrid: List[List[int]]
            :rtype: int
            """
            m, n = len(obstacleGrid), len(obstacleGrid[0])
            dp = [[0] * n for i in xrange(m)]
            for i in xrange(m):
                for j in xrange(n):
                    if obstacleGrid[i][j] != 1:
                        if i == 0 and j == 0:
                            dp[i][j] = 1
                        elif i == 0:
                            dp[i][j] = dp[i][j-1]
                        elif j == 0:
                            dp[i][j] = dp[i-1][j]                        
                        else:
                            dp[i][j] = dp[i-1][j] + dp[i][j-1]
                            
            return dp[-1][-1]  

    C++:

    class Solution {
    public:
        int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
            int m = obstacleGrid.size() , n = obstacleGrid[0].size();
            vector<vector<int>> dp(m+1,vector<int>(n+1,0));
            dp[0][1] = 1;
            for(int i = 1 ; i <= m ; ++i)
                for(int j = 1 ; j <= n ; ++j)
                    if(!obstacleGrid[i-1][j-1])
                        dp[i][j] = dp[i-1][j]+dp[i][j-1];
            return dp[m][n];
        }
    };
    

      

     

    类似题目:

  • 相关阅读:
    Nature:肿瘤转移后的基因组特征
    Nature | 生物体可以从头产生新基因
    一文读懂:DNA甲基化的作用及各种高通量检测方法比较
    Nature | 新技术scSLAM-seq可在单细胞水平揭示转录动态变化的核心特征
    一文读懂长非编码RNA(lncRNA)的分类、功能及测序鉴定方法
    Science重磅 | 新技术Slide-seq能以高空间分辨率测量全基因组的表达情况
    Science综述 | 用单细胞基因组学将人类细胞表型匹配到基因型
    Nature Methods | 新软件SAVER-X可对单细胞转录组学数据进行有效降噪
    Circular RNA的产生机制、功能及RNA-seq数据鉴定方法
    一文搞懂基因融合(gene fusion)的定义、产生机制及鉴定方法
  • 原文地址:https://www.cnblogs.com/lightwindy/p/8469303.html
Copyright © 2011-2022 走看看