[LeetCode]Unique Paths

题目：

从左上角到右下角的所有可能路径。

思路1：

回溯法去递归遍历所有的路径，但是复杂度太大，无法通过。checkPath方法实现

动态规划法，从左上角到每一格的路径数与它的上面一格和左边一格的路径和；

N(m,n)=N(m-1,n)+N(m,n-1)；

注意：第一行和第一列的特殊情况。

package com.example.medium;

/**
 * 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.
 * @author FuPing
 *
 */
public class UniquePaths {
    /**
     * 回溯法
     * @param m 行边界
     * @param n 列边界
     * @param i 当前位置的行坐标
     * @param j 当前位置的列坐标
     * @return 可能的道路数量
     * 时间超过了,20*15的规模就需要7870ms的时间
     */
    private int checkPath(int m,int n,int i,int j){
        if(i == m && j == n)return 1;//到达最后一格
        int roads = 0;
        if(i < m) roads += checkPath(m,n,i+1,j);//向左
        if(j < n) roads += checkPath(m,n,i,j+1);//向下
        return roads;
    }
    /**
     * 动态规划 N(m,n)=N(m-1,n)+N(m,n-1)
     * @param m
     * @param n
     * @return
     */
    private int uniquePaths(int m, int n) {
        //return checkPath(m,n,1,1);
        int roadNums[][] = new int[m][n];
        roadNums[0][0] = 1;
        int i = 0,j = 0;
        for(i = 1;i < m;i++)roadNums[i][0] = roadNums[0][0];//第一行
        for(j = 1;j < n;j++)roadNums[0][j] = roadNums[0][0];//第一列
        i = 1;
        j = 1;
        while(i < m && j < n){
            roadNums[i][j] = roadNums[i][j - 1] + roadNums[i - 1][j];
            j++;
            if(j == n){
                i++;
                if(i == m)break;
                j = 1;
            }
        }
        return roadNums[m - 1][n - 1];
    }

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        long startTime = System.currentTimeMillis();
        System.out.println(new UniquePaths().uniquePaths(20, 50));
        long endTime = System.currentTimeMillis();
        System.out.println("程序运行时间:"+(endTime-startTime) + "ms");

    }

}