LeetCode 64 — Minimum Path Sum（C++ Java Python）

题目：http://oj.leetcode.com/problems/minimum-path-sum/

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.

题目翻译：

给定填充满非负数的m x n网格，找到从左上角到右下角的一条路径，使沿该路径的所有数字的总和最小。

注意：在任何时间点只能向下或向右移动。

分析：

        类似于Unique Paths。

C++实现：

class Solution {
public:
int minPathSum(vector<vector<int> > &grid) {
int m = grid.size();
int n = grid[0].size();

int paths[m]
;

paths[0][0] = grid[0][0];

for(int i = 1; i < m; ++i)
{
paths[i][0] = paths[i - 1][0] + grid[i][0];
}

for(int j = 1; j < n; ++j)
{
paths[0][j] = paths[0][j - 1] + grid[0][j];
}

for(int i = 1; i < m; ++i)
for(int j = 1; j < n; ++j)
{
paths[i][j] = std::min(paths[i - 1][j], paths[i][j - 1])
+ grid[i][j];
}

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

Java实现：

public class Solution {
public int minPathSum(int[][] grid) {
int m = grid.length;
int n = grid[0].length;

int[][] paths = new int[m]
;

paths[0][0] = grid[0][0];

for (int i = 1; i < m; ++i) {
paths[i][0] = paths[i - 1][0] + grid[i][0];
}

for (int j = 1; j < n; ++j) {
paths[0][j] = paths[0][j - 1] + grid[0][j];
}

for (int i = 1; i < m; ++i)
for (int j = 1; j < n; ++j) {
paths[i][j] = Math.min(paths[i - 1][j], paths[i][j - 1])
+ grid[i][j];
}

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

Python实现：

class Solution:
# @param grid, a list of lists of integers
# @return an integer
def minPathSum(self, grid):
m = len(grid)
n = len(grid[0])

paths = [[] for i in range(m)]

paths[0].append(grid[0][0])

for i in range(1, m):
paths[i].append(paths[i - 1][0] + grid[i][0])

for j in range(1, n):
paths[0].append(paths[0][j - 1] + grid[0][j])

for i in range(1, m):
for j in range(1, n):
paths[i].append(min(paths[i][j - 1], paths[i - 1][j])
+ grid[i][j])

return paths[m - 1][n - 1]

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