# [LeetCode] 64. Minimum Path Sum 最小路径和

2021年09月15日 阅读数：2

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.html

Note: You can only move either down or right at any point in time.java

Example:python

Input:
[
[1,3,1],
[1,5,1],
[4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

State: dp[i][j]，表示从(0, 0)到(i, j)最小的路径和code

Function: dp[i][j] = min(dp[i][j-1], dp[i-1][j]) + grid[i][j]htm

Initialize: dp[0][0] = grid[0][0]blog

Return: dp[m - 1][n - 1] 游戏

Java:get

public class Solution {
public int minPathSum(int[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0) {
return 0;
}

int M = grid.length;
int N = grid[0].length;
int[][] dp = new int[M][N];

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 i = 1; i < N; i++) {
dp[0][i] = dp[0][i - 1] + grid[0][i];
}

for (int i = 1; i < M; i++) {
for (int j = 1; j < N; j++) {
dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
}
}

return dp[M - 1][N - 1];
}
}

Python:

class Solution:
# @param grid, a list of lists of integers
# @return an integer
def minPathSum(self, grid):
m = len(grid); n = len(grid[0])
dp = [[0 for i in range(n)] for j in range(m)]
dp[0][0] = grid[0][0]
for i in range(1, n):
dp[0][i] = dp[0][i-1] + grid[0][i]
for i in range(1, m):
dp[i][0] = dp[i-1][0] + grid[i][0]
for i in range(1, m):
for j in range(1, n):
dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j]
return dp[m-1][n-1]

Python:

class Solution:
# @param grid, a list of lists of integers
# @return an integer
def minPathSum(self, grid):
sum = list(grid[0])
for j in xrange(1, len(grid[0])):
sum[j] = sum[j - 1] + grid[0][j]

for i in xrange(1, len(grid)):
sum[0] += grid[i][0]
for j in xrange(1, len(grid[0])):
sum[j] = min(sum[j - 1], sum[j]) + grid[i][j]

return sum[-1]

Python: wo

class Solution(object):
def minPathSum(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
m, n = len(grid), len(grid[0])
dp = [[0] * n for i in xrange(m)]
for i in xrange(m):
for j in xrange(n):
if i == 0 and j == 0:
dp[i][j] = grid[i][j]
elif i == 0:
dp[i][j] = grid[i][j] + dp[i][j-1]
elif j == 0:
dp[i][j] = grid[i][j] + dp[i-1][j]
else:
dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j]

return dp[-1][-1]

Python:

class Solution:
"""
@param grid: a list of lists of integers.
@return: An integer, minimizes the sum of all numbers along its path
"""
def minPathSum(self, grid):
for i in range(len(grid)):
for j in range(len(grid[0])):
if i == 0 and j > 0:
grid[i][j] += grid[i][j-1]
elif j == 0 and i > 0:
grid[i][j] += grid[i-1][j]
elif i > 0 and j > 0:
grid[i][j] += min(grid[i-1][j], grid[i][j-1])
return grid[len(grid) - 1][len(grid[0]) - 1]

C++:

class Solution {
public:
/**
* @param grid: a list of lists of integers.
* @return: An integer, minimizes the sum of all numbers along its path
*/
int minPathSum(vector<vector<int> > &grid) {
// write your code here
int f[1000][1000];
if (grid.size() == 0 || grid[0].size() == 0)
return 0;
f[0][0] = grid[0][0];
for(int i = 1; i < grid.size(); i++)
f[i][0] = f[i-1][0] + grid[i][0];
for(int i = 1; i < grid[0].size(); i++)
f[0][i] = f[0][i-1] + grid[0][i];
for(int i = 1; i < grid.size(); i++)
for(int j = 1; j < grid[0].size(); j++)
f[i][j] = min(f[i-1][j], f[i][j-1]) + grid[i][j];

return f[grid.size()-1][grid[0].size()-1];
}
};

[LeetCode] 174. Dungeon Game 地牢游戏