64. Minimum Path Sum

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.

 

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 Array Dynamic Programming

 

链接： http://leetcode.com/problems/minimum-path-sum/

题解：

DP，可以in-place，也可以使用滚动数组。

Time Complexity - O(mn)， Space Complexity - O(1)。

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

 

Update:

public class Solution {
    public int minPathSum(int[][] grid) {
        if(grid == null || grid.length == 0)
            return 0;
        int rowLen = grid.length, colLen = grid[0].length;
        
        for(int i = 1; i < rowLen; i++)     //initialize first column
            grid[i][0] += grid[i - 1][0];
        
        for(int j = 1; j < colLen; j++)     //initialize first row
            grid[0][j] += grid[0][j - 1];
        
        int sum = 0;
        
        for(int i = 1; i < rowLen; i++) {
            for(int j = 1; j < colLen; j++) {
                grid[i][j] += Math.min(grid[i - 1][j], grid[i][j - 1]);
            }
        }
        
        return grid[rowLen - 1][colLen - 1];
    }
}

 

二刷：

依然是dp，跟前两道题一样。

Java:

Time Complexity - O(mn)， Space Complexity - O(1)。

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

 

题外话：

1/31/2016

这几天头有点昏，做题速度较慢。其实这一个月都很慢，按照计划应该完成了150题才对，结果到现在为止也才60题，差得太多了。

现在是8点，我在犹豫要不要去吃半亩园。9点钟关门， 开车过去大概20分钟的样子。下午刚跑完步，晚上来一顿估计就白跑了 -____-!! 纠结啊

 

三刷:

Java:

Time Complexity - O(mn)， Space Complexity - O(n)。     m为行数，n为列数

public class Solution {
    public int minPathSum(int[][] grid) {
        if (grid == null || grid.length == 0) return Integer.MIN_VALUE;
        int rowNum = grid.length, colNum = grid[0].length;
        for (int i = 1; i < rowNum; i++) grid[i][0] += grid[i - 1][0];
        for (int j = 1; j < colNum; j++) grid[0][j] += grid[0][j - 1];
        
        for (int i = 1; i < rowNum; i++) {
            for (int j = 1; j < colNum; j++) {
                grid[i][j] += Math.min(grid[i - 1][j], grid[i][j - 1]);
            }
        }
        return grid[rowNum - 1][colNum - 1];
    }
}

 

Update:

使用滚动数组。

Java:

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