0064._minimum_path_sum.md

64. Minimum Path Sum

难度: Medium

刷题内容

原题连接

• https://leetcode.com/problems/minimum-path-sum

内容描述

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.

解题方案

思路 1 - 时间复杂度: O(N^2)- 空间复杂度: O(N^2)******

• 经典的动态规划问题，和：72. 编辑距离 类似

dp[i][j]代表到达点(i, j)所需要的最短路径和

beats 94.63%

class Solution(object):
    def minPathSum(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        if not grid or len(grid) == 0:
            return 0
        
        row = len(grid)
        col = len(grid[0]) if row else 0
        
        dp = [[grid[i][j] for j in range(col)] for i in range(row)]
        
        for i in range(1, col):
            dp[0][i] += dp[0][i-1]
        
        for i in range(1, row):
            dp[i][0] += dp[i-1][0]
            
        for i in range(1, row):
            for j in range(1, col):
                dp[i][j] += min(dp[i-1][j], dp[i][j-1])
                
        return dp[-1][-1]

思路 2 - 时间复杂度: O(N^2)- 空间复杂度: O(1)******

其实我们根本不需要额外的空间

beats 94.63%

class Solution(object):
    def minPathSum(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        if not grid or len(grid) == 0:
            return 0
        
        row = len(grid)
        col = len(grid[0]) if row else 0
        
        for i in range(1, col):
            grid[0][i] += grid[0][i-1]
        
        for i in range(1, row):
            grid[i][0] += grid[i-1][0]
            
        for i in range(1, row):
            for j in range(1, col):
                grid[i][j] += min(grid[i-1][j], grid[i][j-1])
                
        return grid[-1][-1]