64. Minimum Path Sum**
https://leetcode.com/problems/minimum-path-sum/
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.
C++ 实现 1
动态规划. 使用 dp[i][j]
表示到达 grid[i][j]
所需要的最小 cost. (将 grid
中所有元素作为行走所需要的 cost). 为了方便做, 这里设置 dp
的大小为 (m + 1) x (n + 1)
.
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<vector<int>> dp(m + 1, vector<int>(n + 1, INT32_MAX));
for (int i = 1; i <= m; ++ i) {
for (int j = 1; j <= n; ++ j) {
if (i == 1 && j == 1) dp[i][j] = grid[i - 1][j - 1];
else dp[i][j] = grid[i - 1][j - 1] + std::min(dp[i - 1][j], dp[i][j - 1]);
}
}
return dp[m][n];
}
};
C++ 实现 2
两年前的代码.
class Solution {
public:
int minPathSum(vector<vector<int>> &grid) {
if (grid.empty())
return 0;
int row = grid.size();
int col = grid[0].size();
vector<vector<int>> dp = vector<vector<int>>(row, vector<int>(col, 0));
dp[0][0] = grid[0][0];
for (int i = 1; i < col; ++i)
dp[0][i] = dp[0][i - 1] + grid[0][i];
for (int i = 1; i < row; ++i)
dp[i][0] = dp[i - 1][0] + grid[i][0];
for (int i = 1; i < row; ++i) {
for (int j = 1; j < col; ++j) {
dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
}
}
return dp[row - 1][col - 1];
}
};