标签：style blog http color os strong ar for art
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).
How many possible unique paths are there?
Above is a 3 x 7 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
分析：此题第一眼想到就是DFS分别从右边格子和下边的格子各自探测然后得到结果
1 public int search(int m,int n,int cur_x,int cur_y){ 2 if(cur_x == m-1 && cur_y == n-1){ 3 return 1; 4 } 5 if(cur_x > m-1 || cur_y > n-1){ 6 return 0; 7 } 8 return search(m,n,cur_x,cur_y+1) + search(m,n,cur_x+1,cur_y); 9 }
提交， 超时，我擦
再想想，递归操作因为有回溯，会有重复探测重复计算的问题，效率低
换个写法
1 public int searchNew(int m,int n){ 2 if(m == 1 || n == 1){ 3 return 1; 4 } 5 return searchNew(m-1,n) + searchNew(m,n-1); 6
还是递归，只不过简洁了点，但是还是超时（递归没搞清楚啊，递归必然是有重复计算的问题）
那就DP填表吧，避免重复计算，分别从1*1的矩阵一直算到m*n，开个m*n的二维数组轻松搞定
但是---忽然想到这种只需要前一步结果的填表其实不需要矩阵，省点空间，开个一维数组保存上一步的结果就行了
1 public int uniquePathsNew(int m,int n){ 2 int [] a = new int[n]; 3 for(int i = 0;i<n;i++){ 4 a[i] = 1; 5 } 6 for(int i=1;i<m;i++){ 7 for(int j = 1;j< n;j++){ 8 a[j] = a[j] + a[j-1]; 9 } 10 } 11 return a[n-1]; 12 }
轻松AC
标签：style blog http color os strong ar for art
原文地址：http://www.cnblogs.com/dijkstra-c/p/3951788.html
