Problem: 64. 最小路径和

思路

👨‍🏫 路飞

💖 朴素版

⏰ 时间复杂度: $O(nm)$
🌎 空间复杂度: $O(nm)$

class Solution {
	public int minPathSum(int[][] grid)
	{
		int n = grid.length;
		if (n == 0)
			return 0;
		int m = grid[0].length;
		int[][] f = new int[n][m];
        f[0][0] = grid[0][0];
		for (int i = 1; i < m; i++)
			f[0][i] = f[0][i - 1] + grid[0][i];
		for (int i = 1; i < n; i++)
			f[i][0] = f[i - 1][0] + grid[i][0];
		for (int i = 1; i < n; i++)
			for (int j = 1; j < m; j++)
				f[i][j] = Math.min(f[i - 1][j], f[i][j - 1]) + grid[i][j];

		return f[n - 1][m - 1];
	}
}

💖 空间优化版

⏰ 时间复杂度: $O(nm)$
🌎 空间复杂度: $O(1)$

class Solution {
    public int minPathSum(int[][] grid) {
        for(int i = 0; i < grid.length; i++) {
            for(int j = 0; j < grid[0].length; j++) {
                if(i == 0 && j == 0) continue;//不处理
                else if(i == 0)  grid[i][j] = grid[i][j - 1] + grid[i][j];//第一行的，只能从左边过来
                else if(j == 0)  grid[i][j] = grid[i - 1][j] + grid[i][j];//第一列的，只能从上面过来
                else grid[i][j] = Math.min(grid[i - 1][j], grid[i][j - 1]) + grid[i][j];//从上面或者左边较小的地方过来
            }
        }
        return grid[grid.length - 1][grid[0].length - 1];
    }
}