【LeetCode】Unique Paths 解题报告
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【题目】
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).
Note: m and n will be at most 100.
【解析】题意:数组 A[m][n] ,从 A[0][0] 到 A[m-1][n-1] 有多少条路径。
对动态规划不熟的同学,可以借这个例子由浅入深理解一下。
二维数组实现:
public class Solution { public int uniquePaths(int m, int n) { // DP with 2 dimensions array int[][] a = new int[m][n]; for (int i = 0; i < m; i++) { a[i][0] = 1; } for (int i = 0; i < n; i++) { a[0][i] = 1; } for (int i = 1; i < m; i++) { for (int j = 1; j < n; j++) { a[i][j] = a[i-1][j] + a[i][j-1]; } } return a[m-1][n-1]; }}
一维数组实现:
public class Solution { public int uniquePaths(int m, int n) { // DP with 1 dimension array int[] a = new int[n]; for (int j = 0; j < n; j++) { a[j] = 1; } for (int i = 1; i < m; i++) { for (int j = 1; j < n; j++) { a[j] += a[j-1]; } } return a[n-1]; }}
一维数组升级版:
public class Solution { public int uniquePaths(int m, int n) { // Advanced: DP with 1 dimension array int row = Math.min(m, n); int col = Math.max(m, n); int[] a = new int[col]; for (int j = 0; j < col; j++) { a[j] = 1; } for (int i = 1; i < row; i++) { a[i] *= 2; for (int j = i+1; j < col; j++) { a[j] += a[j-1]; } } return a[col-1]; }}
这道题还有一个延伸版本:【LeetCode】Unique Paths II 解题报告
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