leetcode题解-64. Minimum Path Sum

题目：

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.

其实本题和之前的Unique Paths思路是一样的，我们按照列进行遍历即可。每一个值更新为到该点的最小路径之和。代码入下：

    public int minPathSum(int[][] grid) {        for(int i=0; i<grid[0].length; i++)            for(int j=0; j<grid.length; j++){                if(i == 0 && j != 0)                    grid[j][i] += grid[j-1][i];                else if(j == 0 && i != 0)                    grid[j][i] += grid[j][i-1];                else                    grid[j][i] += Math.min(grid[j-1][i], grid[j][i-1]);            }        return grid[grid.length-1][grid[0].length-1];    }

此外我们还可以使用dp算法，使用额外的存储空间来保存路径和信息，这种方法可以击败98%的用户，代码入下：

    public int minPathSum1(int[][] grid) {        int[][] memo = new int[grid.length][grid[0].length];        return minPathSumHelper(grid, 0, 0, memo);    }    public int minPathSumHelper(int[][] grid, int row, int col, int[][] memo) {        if(row == grid.length-1 && col == grid[0].length-1) return grid[row][col];        if(memo[row][col] != 0) return memo[row][col];        int rowInc = Integer.MAX_VALUE, colInc = Integer.MAX_VALUE;        if(row < grid.length-1) rowInc = minPathSumHelper(grid, row+1, col, memo);        if(col < grid[0].length-1) colInc = minPathSumHelper(grid, row, col+1, memo);        memo[row][col] = Math.min(rowInc, colInc) + grid[row][col];        return memo[row][col];    }