64. Minimum Path Sum

Problem:

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.

思路：
利用DP算法，到点(i,j)的最短路径和由公式：S[i][j] = min(S[i-1][j], S[i][j-1]) + grid[i][j]确定。注意到达边界上的点只有一条路径（只能一直向右或者一直向下），故边界上的点(i,0)与(0,j)最短路径由下面的公式确定：S[i][0] = S[i-1][0] + grid[i][0]，S[0][j] = S[0][j-1] + grid[0][j]。

Solution:

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>> S(m, vector<int>(n, grid[0][0]));
    
    for (int i = 1; i < m; i++)   S[i][0] = S[i-1][0] + grid[i][0];
    for (int j = 1; j < n; j++)   S[0][j] = S[0][j-1] + grid[0][j];
    for (int i = 1; i < m; i++) {
        for (int j = 1; j < n; j++) {
            S[i][j] = min(S[i-1][j], S[i][j-1]) + grid[i][j];
        }
    }
    return S[m-1][n-1];
}

性能：
Runtime: 12 ms  Memory Usage: 10.8 MB