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leetcode(64) minimum path sum

admin 11月 29, 2020 0

Description:

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.

解法： 经典的动态规划题，这里给出状态转移公式：

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dp[0][0] = grid[0][0];
//第一行
dp[i][j] = dp[i][j - 1] + grid[i][j];
//第一列
dp[i][j] = dp[i - 1][j] + grid[i][j];
//其他
dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];

具体的代码如下：

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

    public int min(int a, int b) {
        if (a > b) {
            return b;
        }
        return a;
    }
}

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