Leetcode--62. Unique Paths

62 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?

①很直观的递归公式f(m,n)=f(m-1,n)+f(m,n-1).可以看到中间有重复的计算f(m-1,m-1).遇到递归有重复计算可以考虑用动态规划，或者备忘录来解决。备忘录还要用到递归，因而最好还是选用动态规划。
②本题动态转移方程比较直观，设为dp[m][n]=dp[m][n-1]+dp[m-1][n];考虑优化存储空间。可以看到在求下一次m的时候即dp[m][k]只用到dp[m-1][k]和本次的dp[m][k-1]因而，只需用一维数组即可；
③到计算dp[m][k]时，数组保存的就是dp[m-1][k-1]的数据变为dp[n]=dp[n]+dp[n-1];由于dp[n-1]用的是本次计算的结果，故而计算顺序为从前到后。

    public int uniquePaths(int m, int n) {        if (m > n)            return uniquePaths(n, m);        if (m == 1)            return 1;        int[] dp = new int[n + 1];        Arrays.fill(dp, 1);        for (int i = 1; i < m; ++i) {            for (int j = 2; j <= n; ++j) {                dp[j] += dp[j - 1];            }        }        return dp[n];    }

63 Unique Paths II

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]                    ]// 本题目和上一题目的区别在于初始化的时候有差别，遇到障碍物则后面的均为0；之后在计算时遇到障碍物直接置结果为0，同时列的第一个元素需要同第一行一样考虑下；

public int uniquePathsWithObstacles(int[][] obstacleGrid) {        if (obstacleGrid == null) {            return 0;        }        int m = obstacleGrid.length, n = obstacleGrid[0].length;        int[] dp = new int[n];        if (obstacleGrid[0][0] != 1) {            dp[0] = 1;        }        for (int i = 1; i < n; ++i) {            if (obstacleGrid[0][i] == 1) {                dp[i] = 0;            } else {                dp[i] += dp[i - 1];            }        }        for (int i = 1; i < m; ++i) {            for (int j = 0; j < n; ++j) {                if (obstacleGrid[i][j] == 1) {                    dp[j] = 0;                } else {                    //若为第一个元素并且不是1，则和上次计算结果一致；若不是第一个元素，则直接计算                    if (j != 0) {                        dp[j] += dp[j - 1];                    }                }            }        }        return dp[n - 1];    }