LeetCode刷题【Array】 Minimum Path Sum

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本文探讨了在一个非负数网格中寻找从左上角到右下角的最小路径和问题。提供了三种解决方法：递归超时方法、动态规划方法及优化后的动态规划方法，并附带详细的代码实现。

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.

方法一：Time 超限，考虑优化

public class Solution {
    public int minPathSum(int[][] grid) {
		if(null==grid||grid.length<=0) return 0;
		return pathSum(grid,0,0);
    }
	
	private int pathSum(int[][] grid, int sr,int sc){
		if(sr==grid.length-1)
			return grid[sr][sc]+pathSum(grid,sr,sc+1);
		if(sc==grid[0].length-1)
			return grid[sr][sc]+pathSum(grid,sr+1,sc);
		int right = grid[sr][sc]+pathSum(grid,sr,sc+1);
		int down =grid[sr][sc]+pathSum(grid,sr+1,sc);
		return Math.min(right,down);
	}
}

方法二：动态规划 Runtime: 5 ms

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

方法三：优化方法二，不利用额外空间

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

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

参考：

【1】https://leetcode.com/