# 64. Minimum Path Sum 最小路径和

@TOC

## # 题目描述

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.

## # 解题方法

class Solution:
def minPathSum(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
if not grid or not grid[0]: return 0
m, n = len(grid), len(grid[0])
path = copy.deepcopy(grid)
for i in range(m):
for j in range(n):
if i == 0 and j == 0:
before = 0
elif i == 0:
before = path[i][j-1]
elif j == 0:
before = path[i-1][j]
else:
before = min(path[i-1][j], path[i][j-1])
path[i][j] = before + grid[i][j]
return path[m-1][n-1]

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

C++代码如下：

class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
const int M = grid.size(), N = grid[0].size();
vector<vector<int>> dp(M, vector<int>(N, 0));
dp[0][0] = grid[0][0];
for (int i = 1; i < M; ++i)
dp[i][0] = dp[i - 1][0] + grid[i][0];
for (int j = 0; j < N; ++j)
dp[0][j] = dp[0][j - 1] + grid[0][j];
for (int i = 1; i < M; ++i) {
for (int j = 1; j < N; ++j) {
dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
}
}
return dp[M - 1][N - 1];
}
};

## # 日期

2018 年 9 月 11 日 —— 天好阴啊 2018 年 12 月 29 日 —— 2018年剩余电量不足1%