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Unique Paths
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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.
题意：给定一个 m * n 的网格，一个机器人要从左上角走到右下角，每次只能向下或向右移动一个位置，
问有多少种走法
思路1：dfs暴力枚举
复杂度：超时了... O(2^n)
思路2：记忆化搜索
用一个数组paths[i][j]记录从 (0,0) 到 (m,n)的路径数
思路3：dp
设置状态为f[i][j]，表示从(0,0)到达网格(i,j)的路径数，则状态转移方程为
f[i][j] = f[i - 1][j] + f[i][j - 1]
复杂度：时间O(n^2) 空间 O(n)
//思路1
int uniquePaths(int m, int n){
if(m < 0 || n < 0) return 0;
if(m == 1 && n == 1) return 1;
return uniquePaths(m - 1, n) + uniquePaths(m, n - 1);
}
//思路2
//path[i][j]表示从(0,0)到(i,j）的路径数
int paths[101][101];
int dfs(int m, int n){
if(m < 0 || n < 0) return 0;
if(m == 1 && n == 1) return 1;
if(paths[m][n] >= 0) return paths[m][n];
return paths[m][n] = dfs(m - 1, n) + dfs(m, n - 1);
}
int uniquePaths(int m, int n){
memset(paths, -1, sizeof(paths));
return dfs(m, n);
}
//思路2另一种写法
//path[i][j]表示从(i,j)到(m - 1,n - 1)的路径数
int paths[101][101];
int mm, nn;
int dfs(int x, int y){
if(x >= mm || y >= nn) return 0;
if(x == mm - 1 && y == nn - 1) return 1;
if(paths[x][y] >= 0) return paths[x][y];
return paths[x][y] = dfs(x + 1, y) + dfs(x, y + 1);
}
int uniquePaths(int m, int n){
mm = m, nn = n;
memset(paths, -1, sizeof(paths));
return dfs(0, 0);
}
//思路3 path[i][j] 表示(0, 0) 到(i,j)的路径数
int paths[101][101];
int uniquePaths(int m, int n){
memset(paths, 0, sizeof(paths));
for(int i = 0; i < m; ++i) paths[i][0] = 1;
for(int j = 0; j < n; ++j) paths[0][j] = 1;
for(int i = 1 ; i < m; ++i){
for(int j = 1; j < n; ++j){
paths[i][j] = paths[i - 1][j] + paths[i][j - 1];
}
}
return paths[m - 1][n - 1];
}