# C++实现LeetCode(64.最小路径和)

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

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

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.

```class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
if (grid.empty() || grid[0].empty()) return 0编程客栈;
int m = grid.size(), n = grid[0].size();
vector<vector<int>> dp(m, vector<int>(n));
dp[0][0] = grid[0][0];
for (int i = 1; i < m; ++i) dp[i][0] = grid[i][0] + dp[i - 1][0];
for (int j = 1; j < n; ++j) dp[0][j] = grid[0][j] + dp[0][j - 1];
for (int i = 1; i < m; ++i) {
for (int j = 1; j < n; ++j) {
dp[i][j] = grid[i][j] + min(dp[i - 1][j], dp[i][j - 1]);
}
}
return dp[m - 1][n - 1];
}
};```
bxUGheeGf

```class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
if (grid.empty() || grid[0].empty()) return 0;
int m = grid.size(), n = grid[0].size();
vector<int> dp(n, INT_MAX);
dp[0] = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (j == 0) dp[j] += grid[i][j];
else dp[j] = grid[i][j] + min(dp[j], dp[j - 1]);
}
}
return dp[n - 1];
}
};```

```class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
if (grid.empty() || grid[0].empty()) return 0;
for (int i = 0; i < grid.size(); ++i) {
for (int j = 0; j < grid[i].size(); ++j) {
if (i == 0 && j == 0) continue;
if (i == 0) grid[0][j] += grid[0][j - 1];
else if (j == 0) grid[i][0] += grid[i - 1][0];
else grid[i][j] += min(grid[i - 1][j], grid[i][j - 1]);
}
}
return grid.back().back();
}
};```

```class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
if (grid.empty() || grid[0].empty()) return 0;
for (int i = 0; i < grid.size(); ++i) {
for (int j = 0; j < grid[i].size(); ++j) {
if (i == 0 && j == 0) continue;
int up = (i == 0) ? INT_MAX : grid[i - 1][j];
int left = (j == 0) ? INT_MAX : grid[i][j - 1];
grid[i][j] += min(up, left);
}
}
return grid.back().back();
}
};```