[LeetCode]Unique Paths

题目：Unique Paths

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

思路2：

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");

    }

}

 题目：Unique PathsII

从左上角到右下角的所有可能路径。这次给定了网格数组，当数组值为1，表示不能通过；即设置了障碍。

思路：

仍然用上面动态规划的方法，只是当遇到障碍物时，该网格的值是0。

package com.example.medium;

/**
 * Follow up for "Unique Paths":
 * Now consider if some obstacles are added to the grids. How many unique paths would there be?
 * An obstacle and empty space is marked as 1 and 0 respectively in the grid.
 * For example,
 * There is one obstacle in the middle of a 3x3 grid as illustrated below.
 * [
 *   [0,0,0],
 *   [0,1,0],
 *   [0,0,0]
 * ]
 * The total number of unique paths is 2.
 * Note: m and n will be at most 100.
 * @author FuPing
 *
 */
public class UniquePaths2 {
    /**
     * 动态规划 N(m,n)=N(m-1,n)+N(m,n-1)
     * @param m
     * @param n
     * @return
     */
    private int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if(obstacleGrid[0][0] == 1)return 0;
        int m = obstacleGrid.length;//行数
        int n = obstacleGrid[0].length;//列数
        System.out.println(m + n);
        int roadNums[][] = new int[m][n];
        roadNums[0][0] = 1;//其实节点为1
        int i = 0,j = 0;
        for(i = 1;i < m;i++){//第一行
            if(obstacleGrid[i][0] == 0)roadNums[i][0] = roadNums[i - 1][0];//没有障碍时，等于左边的路径数
        }
        for(j = 1;j < n;j++){//第一列
            if(obstacleGrid[0][j] == 0)roadNums[0][j] = roadNums[0][j - 1];//没有障碍时，等于上边的路径数
        }
        i = 1;
        j = 1;
        while(i < m && j < n){
            if(obstacleGrid[i][j] == 0)
                roadNums[i][j] = roadNums[i][j - 1] + roadNums[i - 1][j];//没有障碍时，动态规划；否则，还是默认值0
            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();
        int matrix[][] = {{0,0,0,0,0,1,0},{0,0,0,1,0,0,0},{1,0,0,1,0,0,0}};
        System.out.println(new UniquePaths2().uniquePathsWithObstacles(matrix));
        long endTime = System.currentTimeMillis();
        System.out.println("程序运行时间:"+(endTime-startTime) + "ms");

    }

}

题目：Minimum Path Sum

给定一个数组，数组中的值表示通过当前位置的路费，找一条从左上到右下路费最少的路线。

思路：

任然可以用动态规划法。

N(i,j)=A(i,j) + min{N(i-1,j),N(i,j-1)}；

package com.example.medium;

/**
 * Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.
 * Note: You can only move either down or right at any point in time.
 * @author FuPing
 *
 */
public class MinPathSum {
    public int minPathSum(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int pathValues[][] = new int[m][n];
        pathValues[0][0] = grid[0][0];
        int i = 0,j = 0,min = 0;
        for(i = 1;i < m;i++){//第一行
            pathValues[i][0] = pathValues[i - 1][0] + grid[i][0];
        }
        for(j = 1;j < n;j++){//第一列
            pathValues[0][j] = pathValues[0][j - 1] + grid[0][j];
        }
        i = 1;
        j = 1;
        while(i < m && j < n){
            min = pathValues[i][j - 1] > pathValues[i - 1][j] ? pathValues[i - 1][j] : pathValues[i][j - 1];//找左边和上边中较小的路径
            pathValues[i][j] = grid[i][j] + min;//更新当前的路费
            j++;
            if(j == n){//当前行遍历完
                i++;
                if(i == m)break;//全部遍历完
                j = 1;
            }
        }
        return pathValues[m - 1][n - 1];
    }

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        long startTime = System.currentTimeMillis();
        int matrix[][] = {{1,2,3,4,5,6,7},{3,4,1,3,5,2,1},{6,2,3,7,4,2,2}};
        System.out.println(new MinPathSum().minPathSum(matrix));
        long endTime = System.currentTimeMillis();
        System.out.println("程序运行时间:"+(endTime-startTime) + "ms");
    }

}