Lintcode 110. Minimum path sum
Title description: Given an m*n grid containing only non-negative integers, find a path from the upper left corner to the lower right corner that can minimize the sum of the numbers.
- You can only move one step down or right at the same time
It belongs to the problem of the minimum value in coordinate dynamic programming:
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) {
int m = grid.size(), n = grid[0].size();
if (0 == m && n == 0) {
return 0;
}
vector<vector<int> > f(m, vector<int>(n));
//1. 起点
f[0][0] = grid[0][0];
//2. 边界情况
for (int i = 1; i < m; ++i) {
f[i][0] = f[i - 1][0] + grid[i][0];
}
for (int j = 1; j < n; ++j) {
f[0][j] = f[0][j - 1] + grid[0][j];
}
//3. 状态转移
for (int i = 1; i < m; ++i) {
for (int j = 1; j < n; ++j) {
f[i][j] = min(f[i - 1][j], f[i][j - 1]) + grid[i][j];
}
}
//4. 终点
return f[m - 1][n - 1];
}
};