标签:value poi rom == one ret [] tom blog
public class Solution { /** * @param n, m: positive integer (1 <= n ,m <= 100) * @return an integer */ public int uniquePaths(int m, int n) { if (m <= 0 || n <= 0) { return 1; } if (m == 1 || n == 1) { return 1; } // write your code here // initialize //f[x][y]: unique path number from (0,0) to (x, y) int[][] f = new int[m][n]; //inital the first value f[0][0] = 0; //initial the top for (int i = 1; i < n; i++) { f[0][i] = 1; //only one path could reach those points } //initial the left for (int i = 1; i < m; i++) { f[i][0] = 1; } //from top to bottom, from left to right for (int i = 1; i < m; i++) { for (int j = 1; j < n; j++) { f[i][j] = f[i - 1][j] + f[i][j - 1]; } } return f[m - 1][n - 1]; } }
Actually it can initial f[0][0] to 1, so that it can avoid a lot of coner checking.
标签:value poi rom == one ret [] tom blog
原文地址:http://www.cnblogs.com/codingEskimo/p/6901824.html