LeetCode 64 Minimum Path Sum

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.
一个二位数组，里边都是非负数，找一条路径从左上角走到右下角，要求路径上的数字之和最小而且你只能往右或者往下走。
这是一个比较simple的DP题目。
思路：假如现在站在右下角那个矩阵元素上，那么你的上一个元素可以从上边来，可以从左边来，只要选择最小的就可以。那么得出状态转移方程：dp[i][j]=matrix[i][j]+min(dp[i-1][j]+dp[i][j-1])。代码如下:

12345678910111213141516171819

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

看discuss时看到一种不用额外空间的方法：

123456789101112131415

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

discuss还有人自称用dijkstra做出来的……表示膜拜