Unique Paths 求在矩阵里从左上走到右下的总方法数@LeetCode

本文探讨了机器人从网格的左上角移动到右下角的不同路径数量，仅能向右或向下移动。对于一个特定的3x7网格为例，详细解释并计算所有可能的独特路径总数。

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package Level2;

/**
 * Unique Paths 
 * 
 * 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. 
 *
 */
public class S62 {

	public static void main(String[] args) {

	}

	public int uniquePaths(int m, int n) {
		// DP数组用来存放到每一个格子时的路径数目
        int[][] cnt = new int[m][n];
        cnt[0][0] = 1;
        // 第一列的格子只有一种到达方式（向下）
        for(int i=0; i<m; i++){
        	cnt[i][0] = 1;
        }
        // 第一行的格子只有一种到达方式（向右）
        for(int i=0; i<n; i++){
        	cnt[0][i] = 1;
        }
        
        for(int i=1; i<m; i++){
        	for(int j=1; j<n; j++){
        		cnt[i][j] = cnt[i-1][j] + cnt[i][j-1];
        	}
        }
        
        return cnt[m-1][n-1];
    }

}

public class Solution {
    public int uniquePaths(int m, int n) {
        int[][] ways = new int[m][n];
        ways[0][0] = 1;
        for(int i=0; i<m; i++){
            ways[i][0] = 1;
        }
        for(int j=0; j<n; j++){
            ways[0][j] = 1;
        }
        for(int i=1; i<m; i++){
            for(int j=1; j<n; j++){
                ways[i][j] = ways[i-1][j] + ways[i][j-1];
            }
        }
        
        return ways[m-1][n-1];
    }
}