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/* * 63. Unique Paths II * 12.20 by Mingyang * 这里是二维DP,需要注意的是i是从1到m,j是从1到n,这里i与j没有联系。 * 然后二维的n m分别是长与宽 */ 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]; }
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原文地址:http://www.cnblogs.com/zmyvszk/p/5472504.html