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Unique Paths II
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来自 <https://leetcode.com/problems/unique-paths-ii/>
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Follow up for "Unique Paths":
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Now consider if some obstacles are added to the grids. How many unique paths would there be?
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An obstacle and empty space is marked as 1 and 0 respectively in the grid.
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For example,
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There is one obstacle in the middle of a 3x3 grid as illustrated below.
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[
? [0,0,0],
? [0,1,0],
? [0,0,0]
]
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The total number of unique paths is 2.
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Note: m and n will be at most 100.
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题目解读:
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上接题目“Unique Paths”
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现在考虑如果某个方格内添加上障碍物,那一共有多少种唯一路径?在数组中的障碍物和空白分别用1和0表示。
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例如:
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在一个3*3的方格中,有一个障碍物,表示如下
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[
? [0,0,0],
? [0,1,0],
? [0,0,0]
]
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总共的唯一路径数为2.
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注:m 和n的最大值为100.
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解析:
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这个题和Unique Paths一样,都属于动态规划的问题,但此问题中添加了障碍物,到达障碍物下端down或右端right的路径数为down左端的路径数和right上端的路径数。如果把障碍物所在的数组元素值设为0,则就能变成河unique paths的问题相同。
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Java代码:
public class Solution { public int uniquePathsWithObstacles(int[][] obstacleGrid) { //行数 int rows = obstacleGrid.length; //列数 int cols = obstacleGrid[0].length; //如果起始位置被堵死,则没有到达目的地的路径 if(obstacleGrid[0][0]==1) return 0; int value = 1; for(int i=0; i<cols; i++) { if(obstacleGrid[0][i] == 1) value = 0; obstacleGrid[0][i] = value; } //如果第一行或第一列的某个数组元素为1,则到达其后续的行和列中的路径为0种 value =1; for(int i=1; i<rows; i++) { if(obstacleGrid[i][0] == 1) value = 0; obstacleGrid[i][0] = value; } //到达空格i,j的路径等于到达其上方和其左方路径之和 for(int i=1; i<rows; i++) { for(int j=1; j<cols; j++) { if(obstacleGrid[i][j] != 1) obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1]; else obstacleGrid[i][j]=0; } } return obstacleGrid[rows-1][cols-1]; } }
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算法性能:
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原文:http://972459637-qq-com.iteye.com/blog/2243177