Question:

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.

Example:

Input:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

Thinking:

  • Method:
    • 参考62， 63对dp的使用
class Solution {
    public int minPathSum(int[][] grid) {
        if(grid == null || grid.length == 0) return 0;
        int height = grid.length;
        int width = grid[0].length;
        int[][] dp = new int[height][width];
        dp[0][0] = grid[0][0];
        for(int i = 1; i < height; i++)
            dp[i][0] = dp[i - 1][0] + grid[i][0];
        for(int i = 1; i < width; i++)
            dp[0][i] = dp[0][i - 1] + grid[0][i];
        for(int i = 1; i < height; i++){
            for(int j = 1; j < width; j++){
                dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
            }
        }
        return dp[height - 1][width - 1];
    }
}

二刷

思路还是和上一道题类似，我看了一下我的解法似乎和一刷的时候是一样。

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