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.
一刷
題解：
dp的經(jīng)典問題营袜，每次向右或向下走一步变汪。第一行或者第一列走到頭只有一種方法虑凛，所以初始化為1廉赔，轉(zhuǎn)移方程是dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
思路一，構(gòu)造m*n的矩陣保存dp[i][j]
Time complexity O(mn), space complexity O(mn)
public class Solution {
public int uniquePaths(int m, int n) {
int[][] dp = new int[m][n];
for(int i=0; i<m; i++){
for(int j=0; j<n; j++){
if(i == 0||j==0) dp[i][j] = 1;
else dp[i][j] = dp[i-1][j] + dp[i][j-1];
}
}
return dp[m-1][n-1];
}
}
思路二，由于dp[i][j] = dp[j][i]是對稱的，考慮利用這個特點來save space
Time complexity O(mn), space complexity O(n)
public class Solution {
public int uniquePaths(int m, int n) {
if(m<0 || n<0) return 0;
int[] dp = new int[n];
for(int i=0; i<n; i++){
dp[i] = 1;
}
for(int i=1; i<m; i++){
for(int j=1; j<n; j++){
dp[j] = dp[j] + dp[j-1]; //dp[j] represent dp[i-1][j], dp[j-1] represent dp[i][j-1]
}
}
return dp[n-1];
}
}
